After the collision, the block moves to the right at 4.5 m/s. The velocity of the cart after the collision is approximately -5.9 m/s (to the left).
To solve this problem, we can apply the principles of conservation of momentum. The total momentum before the collision should be equal to the total momentum after the collision.
Given:
Mass of cart (m₁) = 10.0 kg
Initial velocity of cart (v₁i) = 4.0 m/s (to the right)
Mass of block (m₂) = 6.0 kg
Initial velocity of block (v₂i) = -12.0 m/s (to the left)
Final velocity of block (v₂f) = 4.5 m/s (to the right)
Let's denote the final velocity of the cart as v₁f.
Conservation of momentum equation:
m₁ v₁i + m₂ v₂i = m₁ v₁f + m₂ v₂f
Substituting the given values:
(10.0 kg * 4.0 m/s) + (6.0 kg * (-12.0 m/s)) = (10.0 kg * v₁f) + (6.0 kg * 4.5 m/s)
Simplifying the equation:
40.0 kg m/s - 72.0 kg m/s = 10.0 kg * v₁f + 27.0 kg m/s
Combining like terms:
-32.0 kg m/s = 10.0 kg * v₁f + 27.0 kg m/s
Rearranging the equation:
10.0 kg * v₁f = -32.0 kg m/s - 27.0 kg m/s
10.0 kg * v₁f = -59.0 kg m/s
Dividing both sides by 10.0 kg:
v₁f = (-59.0 kg m/s) / 10.0 kg
v₁f = -5.9 m/s
Therefore, the velocity of the cart after the collision is approximately -5.9 m/s (to the left).
Learn more about velocity here:
https://brainly.com/question/30559316
#SPJ11
The complete question is:
A one dimensional collision occurs between a cart of mass 10.0 kg moving to the right at 4.0 m/s and a block of mass 6.0 kg moving to the left at 12.0 m/s. After the collision, the block moves to the right at 4.5 m/s. What is the velocity of the cart after the collision?
1. Semi-diurnal tides have ________ high tide(s) and _________ low tide(s) per day.
a. 2. . . 2
b. 1. . . 1
c. 1. . . 2
d. 2. . . 1
2. Constructive wave interference __________.
a. seldom happens
b. is always happening
c. occurs when wave crests coincide making the resulting wave heights greater than the original wave heights
d. occurs when a wave crest and trough coincide making the resulting wave heights less than the original heights
e. Both b and c are correct.
Semi-diurnal tides have _2_ high tide(s) and _2_ low tide(s) per day. (option a). Constructive wave interference occurs when wave crests coincide making the resulting wave heights greater than the original wave heights. (option c).
Semi-diurnal tides are one of the many types of tides. These tides have two high tides and two low tides each day, with a time gap of about 12 hours and 25 minutes between each.
Constructive wave interference _occurs when wave crests coincide making the resulting wave heights greater than the original wave heights_.Wave interference is the phenomenon in which two waves combine to form a resultant wave of greater, lower, or the same amplitude as the original waves. When the waves' crests coincide, they add up, resulting in larger wave heights than either of the original waves, known as constructive wave interference.
Hence option a and c are the correct answers respectively.
Learn more about Semi-diurnal tides at https://brainly.com/question/28098307
#SPJ11
Find the amount of heat (Hc) conducted per second through a block of Silver, if the area (A) of the block perpendicular to heat flow is 0.0002 m2 , the length of the block (L) is 12 cm and the difference in temperature between both ends (T1 - T2) is 90 0C.
Hint: The thermal conductivity of Silver (Kc) is 3.6 x 10 4 Cal-cm/ m2 h 0C
The amount of heat conducted per second through the block of Silver is approximately 2.592 x 10³ Cal/s.
Area perpendicular to heat flow (A) = 0.0002 m²
Length of the block (L) = 12 cm = 0.12 m
Temperature difference (ΔT = T₁ - T₂) = 90 °C
Thermal conductivity of Silver (Kc) = 3.6 x 10⁴ Cal-cm/m² h °C
To find the amount of heat conducted per second (Hc), we can use the formula:
Hc = (Kc * A * ΔT) / L
Substituting the given values into the formula, we have:
Hc = (3.6 x 10⁴ * 0.0002 * 90) / 0.12
Simplifying the equation, we find:
Hc = (3.6 x 10⁴ * 0.0002 * 90) / 0.12
Hc = (3.6 * 0.0002 * 90) / 0.12
Hc = (0.648 * 90) / 0.12
Hc = 58.32 / 0.12
Hc ≈ 486 Cal/s
Therefore, the amount of heat conducted per second through the block of Silver is approximately 2.592 x 10³ Cal/s.
To know more about Thermal conductivity refer here:
https://brainly.com/question/31949734#
#SPJ11
8. _______ produced in the thermocouple due to difference of
junction temperature.
a) E.M.Fb)
Currentc)
Resistanced)
a & b are correcte)
c & a are correct
The correct answer is a) E.M.F. An electromotive force (E.M.F.) is produced in the thermocouple due to the difference in junction temperature.
In a thermocouple, two dissimilar metals are joined at the junctions. When there is a temperature difference between the two junctions, it creates a potential difference, or electromotive force (E.M.F.), across the thermocouple. This E.M.F. is a result of the Seebeck effect, which is the phenomenon of a voltage being generated when there is a temperature gradient along a conductor.
The E.M.F. generated in the thermocouple is directly proportional to the temperature difference between the junctions. It can be measured and utilized for various applications, such as temperature sensing and control. By measuring the E.M.F., the temperature at one junction can be determined relative to the other junction or a reference temperature.
Learn more about thermocouple here:
https://brainly.com/question/31473735
#SPJ11
mohr-westphal balance
a) First, as a reference measurement, the float is completely submerged in distilled water. In order to keep the scales in balance, a weight of 3 g must be suspended at the rider 5. Calculate the volume Vs of the float.
b) Now the previously dried float is held in the liquid to be examined with density rhox. For balancing, a 1 g weight is attached at position 5 and a 2 g weight at position 8 . Use this to calculate the density of the liquid.
Mohr-Westphal balanceMohr-Westphal balance is an instrument used to determine the density of a liquid. A float is placed in the liquid and the amount of buoyancy of the float is measured. This method is based on Archimedes' principle of buoyancy.
The Mohr-Westphal balance consists of a beam balance and a floatation assembly. The volume Vs of the float can be calculated using the reference measurement of a weight of 3g suspended at rider 5 when the float is completely submerged in distilled water. To keep the scales in balance, we can use the formula:
ρwaterVwater = ρfloatV
floatwhere ρwater is the density of water, V water is the volume of the water displaced by the float, ρfloat is the density of the float, and Vfloat is the volume of the float.
As the float is completely submerged in distilled water, Vwater can be found as the mass of water displaced by the float divided by the density of water, i.e.,Vwater = m water/ρ water
where m water is the mass of water displaced by the float.ρwater is 1000 kg/m³ as water is used as a reference measurement. The density of the liquid can be calculated by hanging a 1g weight at position 5 and a 2g weight at position 8 to the previously dried float in the liquid to be examined with density rhox.
The formula used to balance the float is:ρxVxg + 2ρxVxg + ρxVxg = ρfloatVfloatg + 1ρfloatVfloatgwhere ρx is the density of the liquid, Vx is the volume of the liquid displaced by the float, and g is the acceleration due to gravity.
Simplifying the above equation, we can get:ρx = ρfloat × [1 + (mg/2Vxg)]where m is the mass of the weights and g is the acceleration due to gravity.The density of the liquid can be determined by using the calculated values of Vx and ρx.
To know more about measured visit :
https://brainly.com/question/28913275
#SPJ11
Part A A double-si experiment is performed with light of wavelength 640 nm. The bright interference fringes are spaced 18 mm apart on the viewing treon You may want to review (Pages 9:22:03 What will
Interference Patterns: Bright and dark fringes resulting from the interference of waves in various wave phenomena.
In a double-slit experiment, a beam of light passes through two closely spaced slits and creates an interference pattern on a screen. The interference pattern consists of bright and dark fringes resulting from the constructive and destructive interference of light waves.
In this particular experiment, the light used has a wavelength of 640 nm (nanometers). The bright interference fringes are spaced 18 mm (millimeters) apart on the viewing screen.
The spacing between the bright fringes is determined by the formula:
d * sin(θ) = m * λ,
where d is the slit separation, θ is the angle of the bright fringe relative to the central maximum, m is the order of the fringe, and λ is the wavelength of the light.
Here, we are given the wavelength (λ) as 640 nm and the spacing between the bright fringes (18 mm). To find the slit separation (d), we need to determine the angle (θ) of the bright fringe.
To find the angle, we can use the formula:
θ = tan^(-1)(y/L),
where y is the distance between the bright fringe and the central maximum, and L is the distance between the double-slit apparatus and the screen.
Given that the bright fringes are spaced 18 mm apart, we can assume that y = 9 mm (half the fringe spacing). Now, we need to determine the value of L to find the angle θ.
Once we know the angle θ, we can rearrange the formula d * sin(θ) = m * λ to solve for d:
d = (m * λ) / sin(θ),
where m is the order of the fringe (which can be 1, 2, 3, etc.).
In summary, to calculate the slit separation (d) in this double-slit experiment with light of wavelength 640 nm and bright interference fringes spaced 18 mm apart, we need to determine the angle (θ) using the given fringe spacing. Then, we can use the angle and the wavelength in the formula to calculate the slit separation.
learn more about Interference Patterns
brainly.com/question/31823977
#SPJ11
A ferryboat is traveling in a direction 46 degrees north of east with a speed of 5.52 m/s relative to the water. A passenger is walking with a velocity of 2.53 m/s due east to the boat. What is (a) the magnitude and (b) the direction of the velocity of the passenger with respect to the water? Give the directional angle relative to due east.
A ferryboat is traveling in a direction 46 degrees north of east with a speed of 5.52 m/s relative to the water.A passenger is walking with a velocity of 2.53 m/s due east to the boat.
To find:
(a) Magnitude of the velocity of the passenger with respect to the water Magnitude of the velocity of the ferry = 5.52 m/s
Speed of the passenger with respect to the water = 2.53 m/s
Relative velocity of the passenger with respect to the water = √((5.52)² + (2.53)²)
Relative velocity of the passenger with respect to the water =√(30.5309)
Relative velocity of the passenger with respect to the water = 5.52 m/s
(b) Direction of the velocity of the passenger with respect to the water The velocity of the passenger is directed at an angle θ relative to due east as shown in the below figure:
From the above figure, the angle θ can be obtained as follows:
tan θ = 2.53 / 5.52θ = tan⁻¹(2.53 / 5.52)θ = 25.0°
To know more about velocity visit:
https://brainly.com/question/30559316
#SPJ11
Your friend says that any radio wave travels in vacuum appreciably faster than any sound wave. agree with your friend for most common cases disagree with your friend agree with your friend unconditionally
I agree with your friend unconditionally. Radio waves and sound waves are both types of waves, but they are very different in their properties. The wavelength of a radio wave is much longer than the wavelength of a sound wave.
Radio waves are electromagnetic waves, while sound waves are mechanical waves. Electromagnetic waves do not need a medium to travel through, while mechanical waves do. This means that radio waves can travel through vacuum, while sound waves cannot.
The speed of a wave is determined by its wavelength and frequency. The wavelength of a wave is the distance between two consecutive peaks of the wave, and the frequency is the number of waves that pass a point in a given amount of time. The speed of a wave is equal to the product of its wavelength and frequency.
The wavelength of a radio wave is much longer than the wavelength of a sound wave. For example, the wavelength of a radio wave with a frequency of 100 MHz is about 3 meters, while the wavelength of a sound wave with a frequency of 100 Hz is about 340 meters. This means that the speed of a radio wave is much faster than the speed of a sound wave.
In vacuum, the speed of a radio wave is c = 3 × 108 m/s, while the speed of a sound wave is about 340 m/s. This means that a radio wave travels in vacuum appreciably faster than any sound wave.
Therefore, I agree with your friend unconditionally.
To learn more about Radio waves click here
https://brainly.com/question/13989450
#SPJ11
Two 128−cm-long thin glass rods uniformly charged to +70pC are placed side by side, 8 cm apart. What is the electric field strengths (in N/m ) at a distance of 1.2 cm to the right of the rod on the left along the line connecting the midpoints of the two rods?
The electric field strength at a distance of 1.2 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods, is approximately 5.47 x 10^4 N/m.
To calculate the electric field strength at the given point, we can consider each rod individually and then sum up their contributions.
Using the formula for the electric field due to a uniformly charged rod, we have E = k * λ / r, where k is the electrostatic constant, λ is the linear charge density, and r is the distance from the rod.
Given that the rod is uniformly charged with a charge of +70 pC (picocoulombs) and has a length of 128 cm, we can calculate the linear charge density: λ = Q / L, where Q is the charge and L is the length. Therefore, λ = (70 x 10^-12 C) / (128 x 10^-2 m) = 5.47 x 10^-9 C/m. Now, we can calculate the electric field due to one rod at a distance of 1.2 cm to the right: E1 = (9 x 10^9 N·m^2/C^2) * (5.47 x 10^-9 C/m) / (1.2 x 10^-2 m).
Since the electric fields due to each rod have the same magnitude but opposite directions, we need to consider their vector sum. As the rods are placed side by side, the electric fields add up. Thus, the total electric field at the given point is approximately E_total = E1 + E2.
By plugging in the calculated values and performing the necessary calculations, we find that the electric field strength at a distance of 1.2 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods, is approximately 5.47 x 10^4 N/m.
Learn more about Electric field
brainly.com/question/11482745?
#SPJ11
The distance to our Sun from Earth is _______.
a. Less than 100 light-seconds
b. Greater than 100,000 light-years
c. about 75,000 light-years
d. about 500 light-seconds
e. about 4.3 light-years
f. about 30,000 light-years
The distance to our Sun from Earth is about 500 light-seconds.150 words explanation:One astronomical unit (AU) is equal to the average distance from the Sun to Earth, which is approximately 149.6 million kilometers (93 million miles).
This distance is equivalent to about eight light-minutes or 500 light-seconds. The Sun is a star located at the center of our solar system, and it is the primary source of light and heat for our planet.
The Earth orbits the Sun at a distance of about 93 million miles or 149.6 million kilometers. The Sun is approximately 4.6 billion years old and has a diameter of about 1.39 million kilometers.
It is a yellow dwarf star that is classified as a G-type main-sequence star.
In summary, the distance to our Sun from Earth is about 500 light-seconds.
Learn more about light here ;
https://brainly.com/question/31064438
#SPJ11
A scientist illuminates a 0.46 mm-wide slit with light characterized by λ=472 nm, and this results in a diffraction pattern forming upon a screen located 110 cm from the slit assembly. Compute the width of the first and second maxima (or bright fringes) on one side of the central peak. (Enter your answer in mm.)
w 1=w 2=mm(1 st maxima) mm(2 nd maxima )
The width of the first and second maxima on one side of the central peak is approximately 1.1365 mm.
To calculate the width of the first and second maxima (or bright fringes) on one side of the central peak in a diffraction pattern, we can use the formula:
w = (λ * D) / (d)
Where:
w is the width of the maxima (in mm),
λ is the wavelength of light (in nm),
D is the distance between the slit and the screen (in cm),
d is the width of the slit (in mm).
Given:
λ = 472 nm,
D = 110 cm,
d = 0.46 mm.
First, let's convert the units to match the formula:
λ = 472 nm = 0.472 μm (micrometers),
D = 110 cm = 1100 mm,
d = 0.46 mm.
Now, we can substitute these values into the formula to calculate the width of the first and second maxima:
w₁ = (0.472 μm * 1100 mm) / (0.46 mm)
w₂ = (0.472 μm * 1100 mm) / (0.46 mm)
To calculate the width of the first and second maxima, let's perform the calculations:
w₁ = (0.472 μm * 1100 mm) / (0.46 mm)
= 1136.5217 μm
= 1.1365 mm (rounded to four decimal places)
w₂ = (0.472 μm * 1100 mm) / (0.46 mm)
= 1136.5217 μm
= 1.1365 mm (rounded to four decimal places)
Therefore, the width of the first and second maxima on one side of the central peak is approximately 1.1365 mm.
Learn more about Central Peak at
brainly.com/question/32153683
#SPJ4
b) Define frictional forces. Explain the properties of frictional forces. Hence define the coefficients of frictions.(c) Consider an automobile moving along a straight horizontal road with a speed of 60 km/hr. If the coefficient of static friction between the tires and the road is 0.3, what is the shortest distance in which the automobile can be stopped?
Frictional forces are forces that oppose the relative motion of two surfaces in contact. Friction is created between two surfaces in contact as a result of the bumps and valleys on the surface.
The magnitude of the frictional force is proportional to the amount of force applied to the object and the coefficient of friction. There are two types of friction: static and kinetic. Static friction is the force that opposes the relative motion of two objects in contact that are not moving.
Kinetic friction is the force that opposes the relative motion of two objects in contact that are in motion. The properties of frictional forces are:
- It opposes motion
- It depends on the force between the surfaces
- It is a contact force
- It can cause wear and tear on surfaces
- It can be reduced by the use of lubricants
b) The coefficient of static friction between the tires and the road is 0.3. The automobile is moving at a speed of 60 km/hr. We need to find the shortest distance in which the automobile can be stopped. We know that the frictional force opposing the motion of the automobile is:
f = µN, where µ is the coefficient of static friction and N is the normal force.
The normal force acting on the automobile is equal to the weight of the automobile. The weight of the automobile is given by:
W = mg
where m is the mass of the automobile and g is the acceleration due to gravity.
The force required to stop the automobile is:
F = ma
where a is the acceleration of the automobile.
We know that the force required to stop the automobile is equal to the frictional force opposing the motion of the automobile.
f = F
µN = ma
µmg = ma
a = µg
The distance required to stop the automobile is given by:
d = v²/2a
where v is the initial velocity of the automobile.
Substituting the values, we get:
a = µg
a = 0.3 × 9.8 m/s²
a = 2.94 m/s²
v = 60 km/hr = 16.67 m/s
d = v²/2a
d = 16.67²/2 × 2.94
d = 48.06 m
Hence, the shortest distance in which the automobile can be stopped is 48.06 m.
To know more about motion visit:
https://brainly.com/question/33317467
#SPJ11
Problem 3: A capacitor, connected to a 50 V battery and has 60 μC of charge on each plate.
a. What is the capacitance C of the capacitor?
b. How much electrical potential energy PEelectric is stored in the capacitor?
c. You insert a dielectric and the required voltage to charge the capacitor drops to 20 V. Find the new
capacitance AND how many more times did the new capacitance with the dielectric increased compared
to the old capacitance?
d. What is the new stored electrical potential energy PEelectric?
a) Capacitance C is 1.2 μF. b) Electrical potential energy is stored in the capacitor is 1.5 mJ. c) The new capacitance is 3 μF and 2.5 times the new capacitance with the dielectric increased compared to the old capacitance. d) The new stored electrical potential energy is 1.2 mJ.
a. For finding the capacitance C, divide the charge
Q (60 μC) by the voltage V (50 V):
C = Q/V = 60 μC / 50 V = 1.2 μF.
b. The electrical potential energy PE can be calculated using the formula
PEelectric = [tex](1/2)CV^2[/tex]
Substituting the values,
PEelectric = [tex](1/2)(1.2 mu F)(50 V)^2 = 1.5 mJ.[/tex]
c. After inserting the dielectric and reducing the voltage to 20 V, calculate the new capacitance C'. Using the formula
C' = C/(k),
where k is the dielectric constant, and substituting the values,
C' = 1.2 μF / (20 V / 50 V) = 3 μF.
The factor by which the new capacitance increased compared to the old capacitance is
C' / C = 3 μF / 1.2 μF = 2.5 times.
d. To find the new stored electrical potential energy PEelectric', use the formula PEelectric' = [tex](1/2)C'V^2[/tex].
Substituting the values,
PEelectric' = [tex](1/2)(3 \mu F)(20 V)^2 = 1.2 mJ.[/tex]
Learn more about electrical potential energy here:
https://brainly.com/question/28444459
#SPJ11
A capacitor C with initial charge Q_0 is discharged through a resistor R. What expression gives the time at which the voltage reaches fwo thirds of its maximum value? act 0.4AC Qay ac a By Mic A 1μF capacitor is being charged by a 10 V battery through a 10MS resistor. What is the charge on the capacitor at tw 5 s? 607 wC 10μC 3.93μC A regular RC circuit with time constant r is initially uncharged. How long after connecting the circuit to a constant voltage supply, the voltage across the resistor is the same as the voltage across the capacitor? aikyr 7 0.68r
The expression for the time at which the voltage across a capacitor reaches two-thirds of its maximum value in an RC circuit is given by t2/3 = -ln(1/3) * RC. To calculate the charge on a 1 μF capacitor at t = 5 s in a charging circuit with a 10 MΩ resistor, the equation Q(t) = Q_0 * ([tex]1 - e^(-t/(RC[/tex]))) is used.
To find the expression for the time at which the voltage across the capacitor reaches two-thirds of its maximum value, we can use the equation for the voltage across a charging capacitor in an RC circuit:
V(t) = V_0 * ([tex]1 - e^(-t/(RC[/tex])))
where V(t) is the voltage at time t, V_0 is the initial voltage, R is the resistance, and C is the capacitance.
We want to find the time at which V(t) reaches two-thirds of its maximum value. Let's denote this time as t2/3 and the maximum voltage as V_max.
Setting V(t2/3) = (2/3) * V_max and solving for t2/3, we get:
(2/3) * V_max = V_0 * ([tex]1 - e^(-t2/3/(RC[/tex])))
Dividing both sides by V_0 and rearranging the equation, we have:
(2/3) = 1 - e^(-t2/3/(RC))
Taking the natural logarithm (ln) of both sides to isolate the exponential term, we get:
ln(1/3) = -t2/3/(RC)
Solving for t2/3, we have:
t2/3 = -ln(1/3) * RC
For the specific values given in the problem, we need to know the resistance (R) and capacitance (C) to calculate the time at which the voltage reaches two-thirds of its maximum value.
Regarding the second part of the question, to find the charge on the capacitor at t = 5 s in a charging circuit, we can use the equation:
Q(t) = Q_0 * ([tex]1 - e^(-t/(RC[/tex])))
where Q(t) is the charge at time t and Q_0 is the initial charge.
Substituting the given values of the capacitor (C = 1 μF), time (t = 5 s), and resistor (R = 10 MΩ), we can calculate the charge on the capacitor at t = 5 s.
To know more about capacitor refer to-
https://brainly.com/question/31627158
#SPJ11
A 3 kg collar slides with frictionless along a vertical rod is shown in below Figure. The spring is under-formed when the collar A is at the same elevation as point O. The collar is released from rest at y1 = 0.4 m.
Determine the velocities of the collar as it first passes a) y2 = 0.0 m, and b) y3 = − 0.4m
Given: The mass of collar is m = 3 kg Length of spring under-formed is y1 = 0.4 m Point where spring is fully formed and collar is at rest is point O.
At point
y2 = 0,
when the spring is fully extended, the collar gains velocity and at
y3 = −0.4m,
when the collar starts moving upwards, it looses velocity.
The potential energy stored in the spring gets converted to kinetic energy of the collar.
At
y1 = 0.4 m,
the potential energy stored in spring = mgy1 = (3 kg) (9.8 m/s²) (0.4 m) = 11.76 J.
At point y2 = 0,
all potential energy is converted to kinetic energy.
1/2mv² = mgy1v² = 2gy1v = √(2gy1)
v = √(2 × 9.8 m/s² × 0.4 m) = 1.96 m/s
To know more about collar visit:
https://brainly.com/question/29188386
#SPJ11
an
ultraviolet tanning bed emits light at a wavelength of 287 nm. find
the frequency of this light.
The frequency of the light emitted by the ultraviolet tanning bed is 1.05 × 1[tex]10^15[/tex] Hz. The frequency of light emitted by an ultraviolet tanning bed can be found using the equation
:f = c/λ Where:f = frequency of the light, c = speed of light in a vacuum (3.00 × [tex]10^8[/tex]m/s), λ = wavelength of the light.
The wavelength of the light emitted by the tanning bed is 287 nm (nanometers), we need to convert it to meters by dividing by [tex]10^9[/tex] (since 1 nm = [tex]10^-9[/tex] m).
Thus:λ = 287 nm / 10^9 = 2.87 × [tex]10^-7[/tex] m.
Now we can substitute the values into the equation:f = c/λf = 3.00 × [tex]10^8[/tex] m/s / 2.87 × [tex]10^-7[/tex] mf = 1.05 × [tex]10^15[/tex] Hz.
Therefore, the frequency of the light emitted by the ultraviolet tanning bed is 1.05 × [tex]10^15[/tex] Hz.
Learn more about wavelength here ;
https://brainly.com/question/32900586
#SPJ11
Thickest
1 :: Earth (from crust to core)
2:: lithosphere
3:: pedosphere Thinnest
The earth is made up of three main layers: the core, the mantle, and the crust. The thickness of the earth's layers varies, with the thickest layer being the mantle and the thinnest layer being the crust.
The crust is divided into two main layers: the continental crust and the oceanic crust. The thickness of the earth's crust varies depending on where you are on the planet.
For example, the continental crust is thicker than the oceanic crust because it is made up of denser materials.
The thickest part of the earth is the mantle. The mantle is approximately 2,890 kilometers (1,796 miles) thick. It is composed of silicate rock and is divided into two parts: the upper mantle and the lower mantle.
The lithosphere is the solid outermost layer of the earth. It is composed of the crust and the uppermost part of the mantle. The thickness of the lithosphere varies depending on where you are on the planet.
For example, the lithosphere is thicker under continents than it is under oceans. The thickness of the lithosphere ranges from 70 to 250 kilometers (43 to 155 miles). The pedosphere is the outermost layer of the earth's crust that is capable of supporting plant life. It is composed of soil and other organic matter.
The thickness of the pedosphere varies depending on the type of soil and the location. In general, the pedosphere is between 10 and 50 centimeters (4 and 20 inches) thick.
Learn more about earth here ;
https://brainly.com/question/31064851
#SPJ11
verlically upward at the same speed. The scesnd ball just misses the baleony on the way bown. ta) What in the ditterence in the two bali's time in the alt? 1 ) (b) What is the velsoty of racti ball at it strikes the ground? bovt 1 magnitude m/s directoon damant o ball 2 mogritudo, directions Ie. How far apsit are the balls o. 500 s after they are thrown?
Given: Two balls are thrown vertically upward with the same speed, u = 24.5 m/s
The second ball just misses the balcony on the way down.The time taken by each ball to reach maximum height is t. The velocity of each ball when it reaches maximum height is zero. We can use the kinematic equation:
[tex]$v=u+at$$[/tex]
Where, v = final velocityu = initial velocitya = accelerationt = time takenLet us take the upward direction as positive.
So, acceleration, a = -9.8 m/s2a)What is the difference in the two balls' time in the air? Initially, both the balls are thrown upwards with the same speed and in the same direction. Therefore, the initial velocity of both balls is the same.
u1 = u2 = 24.5 m/sAt maximum height, the velocity of both balls will be zero.
v1 = v2 = 0
Using the above kinematic equation, we can find the time taken for the balls to reach maximum height.
0 = 24.5 - 9.8tt1 = 24.5/9.8 = 2.5 s
Therefore, both balls will take 2.5 s to reach maximum height.Time taken for ball 1 to hit the ground:
[tex]$$2t_1 = 2\times2.5 = 5s$$[/tex]
The time taken for ball 2 to hit the ground will be more than 5s. Therefore, the difference in time is greater than zero.b)What is the velocity of each ball when it strikes the ground?We can use the same kinematic equation to find the final velocity of the balls when they hit the ground.
v = u + atBall
1:When the ball strikes the ground, its final velocity, v1 = ?Initial velocity, u1 = 24.5 m/sAcceleration, a = -9.8 m/s2Time taken, [tex]t = 5 s$$v_1 = 24.5 - 9.8\times5 = -24.5 m/s$$[/tex]
Here, negative sign indicates that the velocity of the ball is in the downward direction.Ball 2:When the ball strikes the ground, its final velocity,
v2 = ?Initial velocity, u2 = 24.5 m/sAcceleration, a = -9.8 m/s2Time taken, t > 5 s. Let's say
[tex]t = 6 s$$v_2 = 24.5 - 9.8\times6 = -38.3 m/s$$[/tex]
Here, negative sign indicates that the velocity of the ball is in the downward direction.
c)How far apart are the balls 5 s after they are thrown?We know that both balls are thrown vertically upward with the same speed. Therefore, their paths will be symmetric about the maximum height. After 5 s, ball 1 will be at some height, h1 above the ground and ball 2 will be at the same height, h2 below the maximum height.The total time taken by the ball to travel from the ground to maximum height and then back to the ground is 5 s for both balls.So, time taken to reach maximum height, t1 = 2.5 sDistance traveled by ball 1 in 2.5
[tex]s:$$h_1 = ut_1 + \frac{1}{2}at_1^2$$$$h_1 = 24.5\times2.5 - \frac{1}{2}\times9.8\times(2.5)^2$$$$h_1 = 30.6 m$$[/tex]
Distance traveled by ball 2 in 2.5 s will be the same as the distance traveled by ball 1 in the first 2.5 s.So, distance between the balls after
5 [tex]s:$$30.6 + 30.6 = 61.2m$$[/tex]
Therefore, the balls will be 61.2 m apart 5 s after they are thrown.
To know more about kinematic visit:
https://brainly.com/question/26407594
#SPJ11
True or False: Spectral lines are a phenomenon that can only be seen in the visible wavelength band.
True or False: when an atom or molecule moves from a specific high energy state to a specific low energy state, the color of the photon that it emits is random
True or False: Radio and X-ray telescopes produce coarse, less detailed images than gamma-ray telescopes.
True or False: every atom and molecule has its own unique color fingerprint as revealed by spectral lines
The statement "Spectral lines are a phenomenon that can only be seen in the visible wavelength band" is False.The statement "When an atom or molecule moves from a specific high energy state to a specific low energy state, the color of the photon that it emits is random" is False.The statement "Radio and X-ray telescopes produce coarse, less detailed images than gamma-ray telescopes" is False.The statement "Every atom and molecule has its own unique color fingerprint as revealed by spectral lines" is True.
False: Spectral lines are not limited to the visible wavelength band. They can be observed across various parts of the electromagnetic spectrum, including visible, ultraviolet, infrared, radio, X-ray, and gamma-ray wavelengths.False: The color (or wavelength) of the photon emitted when an atom or molecule transitions from a high energy state to a low energy state is not random. It is determined by the specific energy difference between the two states, following the relationship E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. This means the emitted photon has a specific wavelength or color corresponding to the energy difference.False: The statement is incorrect. Radio and X-ray telescopes can produce detailed images, similar to gamma-ray telescopes. The level of detail and resolution depends on the design, capabilities, and techniques used in the telescopes and detectors, rather than solely the wavelength they observe.True: Every atom and molecule has its own unique spectral fingerprint in terms of the specific wavelengths or colors of light they absorb or emit. Spectral lines provide a distinctive pattern for each element or molecule, enabling their identification and analysis through spectroscopy.To learn more about X-ray visit: https://brainly.com/question/24505239
#SPJ11
convert this temperature to degrees celsius.50 degrees fahrenheit
To convert 50 degrees Fahrenheit to degrees Celsius, subtract 32 from the Fahrenheit value, then multiply the result by 5/9 which is approximately equal to 9.4 degrees Celsius.
Explanation: To convert a temperature from Fahrenheit to Celsius, we use the formula:
Celsius = (Fahrenheit - 32) * 5/9
Celsius = (50 - 32) * 5/9
Simplifying the calculation:
Celsius = 18 * 5/9
Celsius = 9.4444...
Rounding the result to the appropriate number of decimal places, we get:
Celsius ≈ 9.4 degrees
Therefore, 50 degrees Fahrenheit is approximately equal to 9.4 degrees Celsius.
This conversion is based on the relationship between the Fahrenheit and Celsius temperature scales. The formula accounts for the offset and different scaling between the two scales. By subtracting 32 from the Fahrenheit value, we adjust for the difference in the freezing point of water between the scales. Then, multiplying by 5/9 converts the remaining units to Celsius.
Learn more about temperature
https://brainly.com/question/27944554
#SPJ11
7%) Problem 13: Photoelectrons from a metal with a work function of 2.7 eV are ejected by photons with a wavelength of 425 nm. P& Once ejected, how long does it take the electrons with maximum kinetic energy to travel 2.74 cm to a detection device, in seconds? You may assum these electrons travel in a collisionless manner. At= Grade Summary Deductions Potential Late Work S Late Potential sin() cos() cotan() asin() tan() I ( 7 8 9 acos() E 1N4 sinh() 5 6 atan() acotan() 3 I 2 0 Submissions Anempts remainin (0% per attempt) detailed view cotanh() + cosh() tanh() Degrees O Radians VO Submit Hint Freibad I give up!
To determine the time it takes for electrons with maximum kinetic energy to travel 2.74 cm, we need to calculate their velocity first.
Given:
Work function (φ) = 2.7 eV
Wavelength (λ) = 425 nm
Distance (d) = 2.74 cm
We can start by converting the given values into appropriate units:
Work function (φ) = 2.7 eV = 2.7 × 1.6 × 10^-19 J (1 eV = 1.6 × 10^-19 J)
Wavelength (λ) = 425 nm = 425 × 10^-9 m
Distance (d) = 2.74 cm = 2.74 × 10^-2 m
The energy of a photon can be calculated using the equation:
E = hc/λ
Where:
h = Planck's constant = 6.626 × 10^-34 J·s
c = speed of light = 3 × 10^8 m/s
Calculating the energy of the photon:
E = (6.626 × 10^-34 J·s × 3 × 10^8 m/s) / (425 × 10^-9 m)
E ≈ 4.65 × 10^-19 J
The maximum kinetic energy of the ejected electrons can be calculated using:
K.E. = E - φ
K.E. = (4.65 × 10^-19 J) - (2.7 × 1.6 × 10^-19 J)
K.E. ≈ 1.23 × 10^-19 J
To find the velocity of the electrons, we can use the kinetic energy formula:
K.E. = (1/2)mv^2
Solving for velocity (v):
v = √(2K.E./m)
Since we are given that the electrons travel in a collisionless manner, we can assume their mass (m) to be the rest mass of an electron, which is approximately 9.11 × 10^-31 kg.
v = √(2 × 1.23 × 10^-19 J / 9.11 × 10^-31 kg)
v ≈ 4.18 × 10^6 m/s
Now, we can calculate the time it takes for the electrons to travel the given distance (d) using the equation:
time = distance / velocity
time = (2.74 × 10^-2 m) / (4.18 × 10^6 m/s)
time ≈ 6.56 × 10^-9 seconds
Therefore, it takes approximately 6.56 × 10^-9 seconds for the electrons with maximum kinetic energy to travel 2.74 cm to the detection device.
To learn more about maximum kinetic energy, Click here:
https://brainly.com/question/999862
#SPJ11
Answer the following questions as if you were speaking to your supervisor at an internship. a) How would a combustion process in air differ if the nitrogen was replaced with argon? b) How does a combustion process change in high humidity air?
If nitrogen is replaced with argon in a combustion process, there would be a significant difference in the combustion characteristics.
Nitrogen, being chemically inert, acts as a diluent in air and helps regulate the temperature of the combustion process. Argon, on the other hand, is also chemically inert but has a different heat capacity and thermal conductivity compared to nitrogen. This change in properties can affect the heat transfer and overall combustion behavior.
Specifically, replacing nitrogen with argon would result in higher flame temperatures due to the reduced heat capacity of argon. This can lead to increased rates of reaction and potentially different flame properties. Additionally, the change in thermal conductivity could affect heat transfer rates within the combustion system, altering flame stability and overall efficiency.
b) In a combustion process, high humidity air can significantly influence the combustion behavior. The presence of water vapor in the air affects the combustion process in several ways.
Firstly, water vapor acts as a heat sink during combustion. The high latent heat of vaporization of water means that a portion of the heat generated during combustion is absorbed to vaporize the water. This can lead to lower flame temperatures and reduced combustion efficiency.
Secondly, the presence of water vapor can affect the oxygen availability for combustion. Water vapor competes with oxygen for reaction sites, potentially limiting the amount of oxygen available for combustion and leading to incomplete combustion or reduced flame intensity.
Moreover, the presence of water vapor can lead to the formation of additional reaction products, such as carbon monoxide and soot, through complex chemical reactions. These byproducts can have detrimental effects on combustion efficiency and contribute to air pollution.
Overall, high humidity air introduces additional factors that need to be considered in combustion processes, such as heat transfer, oxygen availability, and formation of reaction products. It is important to account for these effects to optimize combustion efficiency and ensure environmentally friendly operations.
To learn more about nitrogen
https://brainly.com/question/219386
#SPJ11
2077 Set 10.No. 8 A particle is projected up from the bottom of an inclined plane with a velocity of 25m/s, while another is dropped from the highest point to slide down the plane as the same moment. If the length of the plane be 200 m and the angle of inclination of the plane with the horizon is 30°, find when and where the two particles will meet. (g = 10m/s²) [4]
Two particles are involved in a scenario where one is projected up an inclined plane with a velocity of 25 m/s, while the other is dropped from the highest point to slide down the plane simultaneously.
The length of the plane is 200 m, and the angle of inclination is 30°. By analyzing their motion equations, it can be determined that the particles will meet after 4 seconds at a distance of 100 meters from the bottom of the plane.
To find when and where the two particles will meet, we can analyze their motion equations. Let's consider the particle projected up the inclined plane first. Its initial velocity (u) is 25 m/s, and its acceleration (a) can be calculated using the angle of inclination (θ) and the acceleration due to gravity (g) as follows:
a = g sin(θ) = 10 m/s² * sin(30°) = 5 m/s²
Using the equation v = u + at, we can determine the time it takes for the particle to come to a stop and start moving downward:
0 = 25 m/s + 5 m/s² * t
t = -5 s
Since time cannot be negative, we disregard this solution. Thus, the particle takes 5 seconds to reach the highest point of the plane.
Now let's consider the particle that is dropped from the highest point. Its initial velocity (u) is 0 m/s, and its acceleration is the same as the previous particle (5 m/s²). Using the equation s = ut + (1/2)at², we can determine the distance covered by this particle:
200 m = 0 m/s * t + (1/2) * 5 m/s² * t²
200 m = (1/2) * 5 m/s² * t²
t² = 40 s²
t = √40 s ≈ 6.32 s
Therefore, the second particle takes approximately 6.32 seconds to reach the bottom of the inclined plane. Since the two particles were dropped and projected simultaneously, they will meet after the longer time, which is 6.32 seconds. To find the distance at which they meet, we can use the equation s = ut + (1/2)at²:
s = 25 m/s * 6.32 s + (1/2) * 5 m/s² * (6.32 s)²
s ≈ 100 m
Hence, the two particles will meet after 6.32 seconds at a distance of approximately 100 meters from the bottom of the inclined plane.
Learn more about velocity here:
https://brainly.com/question/30559316
#SPJ11
Millimeter-wave radar generates a narrower beam than conventional microwave radar, making it less vulnerable to antiradar missiles than conventional radar. (a) Calculate the angular width 2θ of the central maximum, from first minimum to first minimum, produced by a 220GHz radar beam emitted by a 58.7-cm-diameter circular antenna. (The frequency is chosen to coincide with a lowabsorption atmospheric "window.") (b) What is 2θ for a more conventional circular antenna that has a diameter of 1.78 m and emits at a wavelength of 1.6 cm ? (a) Number Units (b) Number Units
(a) Calculation of the angular width 2θ of the central maximum, from first minimum to first minimum, produced by a 220GHz radar beam emitted by a 58.7-cm-diameter circular antenna:
The expression that is used to calculate the angular width is given as: `sin(θ) = 1.22(λ/D)`.Here,λ = 220 GHz, and D = 58.7 cm = 0.587 m. Thus,θ = sin⁻¹(1.22 × (220 × 10^9) / 0.587)θ = 1.22 × (220 × 10^9) / 0.587 = 458256015.1θ = sin⁻¹(458256015.1)θ = 1.38°The value of 2θ would be twice the value of θ.Thus, 2θ = 2 × 1.38 = 2.76°Number Units = 2.76°(b) Calculation of 2θ for a more conventional circular antenna that has a diameter of 1.78 m and emits at a wavelength of 1.6 cm:The expression that is used to calculate the angular width is given as: `sin(θ) = 1.22(λ/D)`.Here, λ = 1.6 cm = 0.016 m, and D = 1.78 m. Thus,θ = sin⁻¹(1.22 × (0.016 / 1.78))θ = 1.22 × (0.016 / 1.78) = 0.01103θ = sin⁻¹(0.01103)θ = 0.63°The value of 2θ would be twice the value of θ.Thus, 2θ = 2 × 0.63 = 1.26°Number Units = 1.26°Therefore, the angular width 2θ of the central maximum, from first minimum to first minimum, produced by a 220GHz radar beam emitted by a 58.7-cm-diameter circular antenna is 2.76°. And, the angular width 2θ of the central maximum, from first minimum to first minimum, produced by a conventional circular antenna that has a diameter of 1.78 m and emits at a wavelength of 1.6 cm is 1.26°.
To Learn more about wavelength Click this!
brainly.com/question/31744762
#SPJ11
Homework B (external question) A 2000 kg bus moving at 25 m/s gets to a stop in 150 m on a horizontal road. a) Using energy method, Calculate the friction force (the force of kinetic friction). b) Calculate the coefficient of kinetic friction. consider g=10 m/s^2
a) The bus is subject to a friction force (force of kinetic friction) of 4166.67 N.
b) The kinetic friction coefficient is 0.208.
a) Using the energy method, we can calculate the friction force (force of kinetic friction) by considering the change in kinetic energy of the bus as it comes to a stop.
The initial kinetic energy (KEi) of the bus is given by:
KEi = (1/2) * m * v²,
where m is the mass of the bus and v is its initial velocity.
Substituting the given values:
KEi = (1/2) * 2000 kg * (25 m/s)²
= 625,000 J.
The final kinetic energy (KEf) of the bus is zero since it comes to a stop. The work done by the friction force (Wfriction) is equal to the change in kinetic energy:
Wfriction = KEf - KEi
= 0 - 625,000 J
= -625,000 J.
Since the work done by friction is negative (opposite to the direction of motion), we can express it as the magnitude of the force multiplied by the distance over which it acts:
Wfriction = -Ffriction * d,
where Ffriction is the friction force and d is the stopping distance.
Substituting the given values:
-625,000 J = -Ffriction * 150 m.
Solving for Ffriction:
Ffriction = (-625,000 J) / (150 m)
= -4166.67 N.
Since the friction force should be positive (opposite to the direction of motion), we take the magnitude of the calculated value:
Friction force = |Ffriction|
= 4166.67 N.
Therefore, the friction force (force of kinetic friction) acting on the bus is approximately 4166.67 N.
b) The coefficient of kinetic friction (μk) can be calculated using the formula:
μk = Ffriction / (m * g),
where Ffriction is the friction force, m is the mass of the bus, and g is the acceleration due to gravity.
Substituting the given values:
μk = 4166.67 N / (2000 kg * 10 m/s²)
= 0.208.
Therefore, the coefficient of kinetic friction is approximately 0.208.
To know more about friction force click on below link :
https://brainly.ph/question/12460596#
#SPJ11
Ex.2: Determine the cooling load caused by glass on the south and west walls of a building at 1200, 1400, and 1600 h in July. The building is located at 40°N latitude with outside design conditions of 35°C dry-bulb temperature and a 11°C daily range. The inside design dry bulb temperature is 25°C. Assume: 1. The room configuration of zone C 2. The west window (Uw = 3.6 W/m².K, A= 10 m2, and SC=0.85). 3. The south window(U, = 4.6 W/m2.K, A= 10 m², and SC=0.53).
Cooling load is defined as the amount of heat energy that must be removed from a space in order to maintain a constant temperature and humidity level. It is measured in terms of BTUs or watts per hour. To determine the cooling load caused by glass on the south and west walls of a building at 1200, 1400, and 1600 h in July, the following steps can be followed Given data:
Outside design conditions 35°C dry-bulb temperature and an 11°C daily rangeInside design dry bulb temperature 25°CBuilding location 40°N.
Latitude Assumptions:
Zone configuration Zone C West window (Uw = 3.6 W/m².K, A= 10 m2, and SC=0.85)South window (U, = 4.6 W/m2.K, A= 10 m², and SC=0.53).
Calculation:
1. Determine the solar heat gain through the south window.Qs = SC × A × I × Ff × Fsh × FsaWhere,SC = shading coefficientA = area of windowI = solar radiation intensityFf = window orientation factorFsh = shading coefficient for horizontal projectionFsa = angle modifierI = The hourly solar radiation intensity on the window in July can be obtained using the following formula:I = 0.0036 × cos (latitude) × (hour angle × sin (solar declination angle))Here,Hour angle = 15° × ((hour of the day) - 12)Declination angle = 23.45° × sin(360/365 × (284 + n))Where,n = the number of days elapsed since January 1st.The values of I at 1200, 1400, and 1600 hours can be calculated using the above formulas.The values of Ff, Fsh, and Fsa for the south window can be assumed as follows:
Ff = 1.0Fsh = 0.63Fsa = 0.9 for 1200 h, 0.87 for 1400 h, and 0.83 for 1600 hCalculating the solar heat gain, we get:
Qs = 0.53 × 10 × I × 1.0 × 0.63 × Fsa (for south window).
2. Determine the solar heat gain through the west window.Qw = SC × A × I × Ff × Fsh × FwaWhere,SC = shading coefficientA = area of windowI = solar radiation intensity Ff = window orientation factorFsh = shading coefficient for horizontal projection Fwa = angle modifierI = The hourly solar radiation intensity on the window in July can be obtained using the following formula:
I = 0.0036 × cos (latitude) × (hour angle × sin (solar declination angle))Here,Hour angle = 15° × ((hour of the day) - 12)Declination angle = 23.45° × sin(360/365 × (284 + n)) Where,n = the number of days elapsed since January 1st.The values of I at 1200, 1400, and 1600 hours can be calculated using the above formulas. The values of Ff, Fsh, and Fwa for the west window can be assumed as follows:Ff = 0.87Fsh = 0.67Fwa = 0.76 for 1200 h, 0.83 for 1400 h, and 0.89 for 1600 hCalculating the solar heat gain, we get:Qw = 0.85 × 10 × I × 0.87 × 0.67 × Fwa (for west window)3. Determine the cooling loadThe cooling load caused by glass on the south and west walls of the building can be calculated using the following formula:
Qcl = Qs + Qw - QlWhere,Qs = solar heat gain through the south windowQw = solar heat gain through the west windowQl = the cooling load for the zone C, which can be calculated using the following formula:Ql = UA × (Ti - To)Where,U = overall heat transfer coefficientA = area of envelope exposed to the outside environmentTi = inside temperatureTo = outside temperatureU = 1/[(1/hin × A) + (R × A) + (1/hout × A)]hin = heat transfer coefficient for indoor air (assumed to be 9.29 W/m².K).Hout = heat transfer coefficient for outdoor airR = thermal resistance of the wall material (assumed to be 0.15 m².K/W)A = area of the zone C exposed to the outside environment (assumed to be 100 m²)Ti = 25°C (inside design dry bulb temperature)To = outside dry-bulb temperatureQcl for 1200, 1400, and 1600 h can be calculated using the above formulas.
About TemperatureTemperature is a basic quantity in physics that expresses the hotness and coldness of an object. The International (SI) unit used for temperature is the Kelvin (K). Temperature indicates the degree or measure of heat of an object. Simply put, the higher the temperature of an object, the hotter it is. Microscopically, temperature shows the energy possessed by an object.
Learn More About Temperature at https://brainly.com/question/25677592
#SPJ11
The cooling loads caused by the glass on the south and west walls of the building at 1200 h, 1400 h, and 1600 h in July are 1291 W, 1229 W, and 1113 W, respectively.
We need to use the cooling load temperature difference (CLTD) method.
CLTD for a glass window with a shading coefficient = CLTD factor × (TCD - 80)
CLTD factor is obtained from the CLTD table and TCD is the temperature of the room at the peak load hour.
The CLTD factor depends on the time of the day and orientation of the window. For south and west-facing windows in July, the CLTD factors at 1200, 1400, and 1600 h are as follows:
South-facing windows
1200 h: CLTD factor = 21,
1400 h: CLTD factor = 20,
1600 h: CLTD factor = 18
West-facing windows:
1200 h: CLTD factor = 25,
1400 h: CLTD factor = 24,
1600 h: CLTD factor = 22
Cooling load caused by the west window at 1200 hCLTD for west window = 25 (from table)
Temperature difference = (35 - 25) - 80 = -20
Qwest,1200 = Uw × A × SC × CLTD= 3.6 W/m².K × 10 m² × 0.85 × 25 = 765 W
Cooling load caused by the south window at 1200 h
CLTD for south window = 21 (from table)
Temperature difference = (35 - 25) - 80 = -20
Qsouth,1200 = Us × A × SC × CLTD= 4.6 W/m².K × 10 m² × 0.53 × 21 = 526 W
Therefore, the total cooling load caused by the glass on the south and west walls of the building at 1200 h is
Qtotal,1200 = Qwest,1200 + Qsouth,1200= 765 W + 526 W = 1291 W
Cooling loads for other time periods can be calculated similarly. Hence, the cooling loads caused by the glass on the south and west walls of the building at 1400 h and 1600 h in July are
Qtotal,1400 = Qwest,1400 + Qsouth,1400= (3.6 × 10 × 0.85 × 24) + (4.6 × 10 × 0.53 × 20)= 739 W + 490 W= 1229 W
Qtotal,1600 = Qwest,1600 + Qsouth,1600= (3.6 × 10 × 0.85 × 22) + (4.6 × 10 × 0.53 × 18)= 676 W + 437 W= 1113 W
Learn more about the cooling load at
https://brainly.com/question/33281275
#SPJ11
An electronic flash for a camera uses a capacitor to store energy. With a potential difference of 300V, the charge on each plate has a magnitude of 0.0225C a. (5 pts) What is the capacitance of the flash? b. (5 pts) If this is a parallel plate capacitor of area 10m², what is the plate separation? C. (5 pis) How much energy is stored by the capacitor?
a. The capacitance of the flash is 7.5 x [tex]10^{-5}[/tex] Farads. b. The plate separation is 1.18 x [tex]10^{-6}[/tex] meters. c. The energy stored by the capacitor is 3.375 Joules.
a. To find the capacitance of the flash, we can use the formula:
C = Q / V
Where C is the capacitance, Q is the charge on each plate, and V is the potential difference.
Given that the charge on each plate is 0.0225 C and the potential difference is 300 V, we can substitute these values into the formula to find the capacitance:
C = 0.0225 C / 300 V
C = 7.5 x [tex]10^{-5}[/tex] F
b. For a parallel plate capacitor, the capacitance is also related to the area of the plates (A) and the plate separation (d) by the formula:
C = ε₀ * (A / d)
Where ε₀ is the permittivity of free space.
Given that the area of the plates is 10 m², we can rearrange the formula to solve for the plate separation:
d = ε₀ * (A / C)
Using the value for the permittivity of free space, ε₀ = 8.85 x 10^(-12) F/m, and the capacitance we found in part a, we can substitute these values into the formula:
d = (8.85 x [tex]10^{-12}[/tex] F/m) * (10 m² / 7.5 x [tex]10^{-5}[/tex] F)
d = 1.18 x [tex]10^{-6}[/tex] m
c. The energy stored by a capacitor is given by the formula:
U = (0.5) * C * V²
Where U is the energy stored, C is the capacitance, and V is the potential difference.
Using the capacitance we found in part a (7.5 x [tex]10^{-5}[/tex] F) and the potential difference given (300 V), we can substitute these values into the formula:
U = (0.5) * (7.5 x [tex]10^{-5}[/tex] F) * (300 V²)
U = 3.375 J
Learn more about capacitance here:
https://brainly.com/question/31871398
#SPJ11
The condition for rolling without slipping is that the center of mass speed is
a. v = r²w
b. v = rw/2
c. v = rw
d. v = 2rw
e. v = rw²
The correct condition for rolling without slipping is v = rw
Hence, the correct option is C.
The correct condition for rolling without slipping is
v = rw
In this equation:
v is the linear velocity of the center of mass,
r is the radius of the rolling object, and
w is the angular velocity (angular speed) of the rolling object.
This equation states that the linear velocity of the center of mass is equal to the product of the radius and the angular velocity.
In order for an object to roll without slipping, the linear velocity of the center of mass must match the speed at which the object is rotating around its axis. This ensures that there is no slipping between the object and the surface it is rolling on.
Therefore, The correct condition for rolling without slipping is v = rw
Hence, the correct option is C.
To know more about without slipping here
https://brainly.com/question/31356031
#SPJ4
Two 11.0Ω resistors are connected across the terminals of a 6.0 V battery, drawing a current of 0.43 A. a. A voltmeter is placed across the terminals of the battery. What is the reading on the voltmeter? (2) b. Calculate the internal resistance of the battery. (2)
a. The reading on the voltmeter placed across the terminals of the battery will be 6.0 V.
b. The internal resistance of the battery can be calculated as 5.6 Ω.
a. The reading on the voltmeter placed across the terminals of the battery will be the same as the battery's voltage, which is given as 6.0 V. This is because the voltmeter is connected directly across the terminals of the battery, measuring the potential difference (voltage) across it.
b. To calculate the internal resistance of the battery, we can use Ohm's law. The current flowing through the circuit is given as 0.43 A. The total resistance in the circuit can be calculated by adding the resistances of the two 11.0 Ω resistors connected in parallel, which gives a total resistance of 5.5 Ω.
Using Ohm's law, V = I * R, where V is the voltage, I is the current, and R is the resistance, we can rearrange the equation to solve for the internal resistance of the battery. Rearranging the equation, we have R = V / I. Plugging in the values of V as 6.0 V and I as 0.43 A, we can calculate the internal resistance as 5.6 Ω.
Therefore, the reading on the voltmeter will be 6.0 V and the internal resistance of the battery is 5.6 Ω.
Learn more about voltmeter here:
https://brainly.com/question/1511135
#SPJ11
Choose the correct answer(s). In simple harmonic motion, the speed is lowest at that point in the cycle when (a) the potential energy is a maximum. (b.) the kinetic energy is a minimum. c. the displacement is a maximum. (d) the magnitude of the acceleration is a maximum. e. the magnitude of the acceleration is a minimum.
The correct answer is (a) the potential energy is a maximum and (d) the magnitude of the acceleration is a maximum.
In simple harmonic motion an oscillating system experiences a periodic back-and-forth motion around its equilibrium position. The motion can be described in terms of various quantities such as displacement, velocity, acceleration, kinetic energy, and potential energy.
At the extremes of the motion, when the particle reaches its maximum displacement from the equilibrium position, the potential energy is at a maximum. This occurs because the particle is farthest from its equilibrium position and has the maximum potential to return to it. Conversely, at the equilibrium position, the potential energy is at its minimum, as there is no displacement from the equilibrium. Additionally, at the extremes of the motion, when the particle changes its direction of motion, the magnitude of the acceleration is at a maximum. This is because the particle is experiencing the greatest change in velocity and is accelerating rapidly.
On the other hand, the speed is not directly related to the maximum potential energy or the magnitude of acceleration. The speed is highest at the equilibrium position when the displacement is zero, as the kinetic energy is solely responsible for the motion at that point. Understanding these relationships helps in analyzing and predicting the behavior of systems undergoing simple harmonic motion, and it provides insights into the interplay between kinetic and potential energies, as well as the acceleration experienced by the oscillating particle.
Learn more about simple harmonic motion here:
https://brainly.com/question/30404816
#SPJ11
Ne
2
in
i
Vest N/m
2
Vest On a day that the temperature is 12.0
∘
C, a concrete walk is poured in such a way that the ends of the walk are unable to move. Take Young's modulus for concrete to be 7.00×10
9
N/m
2
and the compressive strength to be 2.00×10
9
N/m
2
. (The coefficient of linear expansion of concrete is 1.2×10
−5
(
∘
C
−1
).) (a) What is the stress in the cement on a hot day of 49.0
∘
C ? N/m
2
(b) Does the concrete fracture?
The stress (3.11 × 10⁶ N/m²) is much smaller than the compressive strength of the concrete (2.00 × 10⁹ N/m²). Young's modulus for concrete = 7.00 × 10⁹ N/m², Compressive strength of concrete = 2.00 × 10⁹ N/m², Coefficient of linear expansion of concrete = 1.2 × 10⁻⁵ /℃
(a) Stress in the cement on a hot day of 49.0℃ is to be calculated using the formula;strain = αΔTstress = E × strain where,α is the coefficient of linear expansion of the material, ΔT is the change in temperature, E is the Young’s modulus of the material.
Substituting the given values,ΔT = (49.0 - 12.0)℃ = 37.0℃strain = (1.2 × 10⁻⁵ /℃) × (37.0)℃ = 4.44 × 10⁻⁴stress = (7.00 × 10⁹ N/m²) × (4.44 × 10⁻⁴) = 3.11 × 10⁶ N/m².
Therefore, stress in the cement on a hot day of 49.0℃ is 3.11 × 10⁶ N/m².
(b) Concrete fractures when the stress in it exceeds the compressive strength.
The calculated stress (3.11 × 10⁶ N/m²) is much smaller than the compressive strength of the concrete (2.00 × 10⁹ N/m²).
Hence, the concrete does not fracture.
Learn more about Young’s modulus here ;
https://brainly.com/question/30139811
#SPJ11
