43.DB(decibels)
⇉ Unit conversion
4.3(bels)
4.951NP
4.3(decades)
14.28(octaves)
Interpretations
⇉ Power quantity level
Basic unit dimensions
⇉ ( logarithmic quantity ).
They provided the intensity in decibels for the problem, but they are unsure by what factor to increase it (I) to make the sound loud enough for the elderly person to hear.
Neglect f entirely.
The following equation must be used to convert decibels (dB) to I:
I=(10^(dB/10))*10^-12
Divide the elder person's dB by the younger person's dB after doing this for each dB.
1.8372093023.
Decibels are used to measure sound (dB). A motorcycle engine operating is roughly 95 dB louder than regular conversation, which is around 60 dB louder than a whisper. Your hearing may begin to be harmed if exposed to noise over 70 dB for an extended period of time. Your ears can suffer instant damage from loud noise above 120 dB.
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Maya made this picture to represent a chemical reaction:
Which of the following statements best explains the type of chemical reaction represented by Maya's picture?
A. It is neither a synthesis reaction nor a decomposition reaction because the total mass of the products is less than the total mass of the reactants.
B. It is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.
C.It represents a synthesis reaction because the same atoms are present in the reactants and products.
D. It represents a decomposition reaction because two reactants break apart and form two products.
The statement that best explains the type of chemical reaction represented by Maya's picture is that it is neither a synthesis reaction nor a decomposition reaction because two reactants form two products. That is option B.
What is a chemical reaction?A chemical reaction is the combination of two elements to yield a new product through the formation of bonds.
A chemical reaction is said to be a synthesis reaction when when two different atoms or molecules interact to form a different molecule or compound.
A chemical reaction is said to be a decomposition reaction when one reactant breaks down into two or more products.
Therefore, from the picture, the chemical reaction is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.
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Most towns use a water tower to store water and provide pressure in the pipes that deliver water to customers. The figure below shows a spherical water tank that holds 5.80 105 kg of water when full. Note that the tank is vented to the atmosphere at the top and that the pipe delivering water to customer Smith is a height h = 3.75 m above the level of the pipe delivering water to customer Jones. Determine the gauge pressure of the water at the faucet of each house when the tank if full.
There is a spherical water-filled chamber with a vent on top and underneath it is a pipe 18 m long that leads down to the ground. At ground level a horizontal pipe connects the faucet in the Jones house to the water supply. The faucet in the Smith house is connected to the water supply with a pipe at a height h above ground level.
(a) Jones house
Pa
(b) Smith house
Pa
The effective height of the water for Smith's house will be 24.61m.
How to calculate the height?Based on the information given, the volume of the water in sphere will be:
= 4/3πr³ = (5.80 × 10^5)/1000
= 4.18r³ = 580
r³ = 138.7
r = 5.18m
The effective height of the water will be:
= 18.0 + 2(5.18)
= 28.36
The gauge pressure at Faucet of Jones house will be:
= pgh
= 1000(9.8)(28.36)
= 277.9kPa
The effective height of the water for Smith's house will be:
= 18.0 + 2(5.18) - 3.75
= 24.61m
The gauge pressure at Faucet of Jones house will be:
= 1000 × 9.8 × 24.61
= 241.2kPa
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A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and
400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the
steam is cooled. Determine the compression work if the final state is
a) 1.0 MPa and 250°C and
b) 500 kPa.
c) Also determine the temperature at the final state in part b.
A piston-work cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. compression work is 44.32 KJ,
The amount of labor put into the piston to cause its reciprocating motion is known as the piston work. It is calculated by multiplying the piston's displacement by the net force.
An expanding gas cylinder's force output is transferred by pistons to the crankshaft, which then drives the flywheel's rotation. A reciprocating engine is a device like this.
Piston work is the effort made by the piston to make its reciprocating motion. The piston's displacement is calculated by multiplying it by the net force.
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a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision
The correct answer is 5828.675 J.
Given combined mass 4kg and mass of bullet 150gm=0.150kg.
Total mass= 4+0.150=4.150kg
Velocity=53 m/s
Kinetic energy = [tex]\frac{1}{2} *m*v^{2}[/tex] =0.5*4.150*[tex]53^{2}[/tex] =5828.675 J
Kinetic energyKinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by being subjected to a net force, it accelerates and consequently obtains kinetic energy. A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. Any combination of motions, including translation (or travel along a path from one location to another), rotation about an axis, and vibration, may be used as the type of motion.
A body's translational kinetic energy is equal to [tex]\frac{1}{2} *m*v^{2}[/tex] , or one-half of the product of its mass, m, and square of its velocity, v.
a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision
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a guage is connected to tank in which the pressure of the fluid is 305kpa above atmospheric. if the absolute presssure of the fluid remains unchanged but the gage is in chamber where the air pressure is reduce to a vaccum of 648mmHg what reading in psi will then be observed
The pressure reading in psi observed after the gauge is placed in the chamber with a vacuum of 648 mmHg will be approximately 31.67 psi.
When the gauge is placed in a chamber with a vacuum of 648 mmHg we need to convert the pressure reading to a compatible unit to determine the pressure measurement in PSI.
First, convert the initial pressure of the fluid from kPa to mmHg.
7.5 mmHg is equal to about 1 kPa. So, 305 kPa * 7.5 mmHg/kPa = 2287.5 mmHg is the initial pressure of the fluid in mmHg.
Reducing the original pressure by vacuum pressure.
1639.5 mmHg = new pressure = 2287.5 mmHg - 648 mmHg
Converting the mmHg of the new pressure to the psi unit.
0.01934 psi is equal to approximately 1 mmHg. As a result, the new pressure in psi is equal to 31.67 psi: 1639.5 mmHg * 0.01934 psi/mmHg
Therefore, the pressure reading in psi observed after the gauge is placed in the chamber with a vacuum of 648 mmHg will be approximately 31.67 psi.
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The graph shows the US Department of Labor noise regulation for working without ear protection. A machinist is in an environment where the ambient sound level is of 85 dB, i.e., corresponding to the 8 Hours/day noise level. The machinist likes to listen to music, and plays a Boom Box at an average level of 86.0 dB.
1. Calculate the INCREASE in the sound level from the ambient work environment level (in dB).
2. A student at a rock concert has a seat close to the speaker system, where the average sound intensity corresponds to a sound level of 116 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?
(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.
(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
Increase in the sound levelThe INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;
Increase in sound level = final sound level - original sound level
Increase in sound level = 86 dB - 85 dB = 1 dB
Factor of sound level increasefrom the graph at 1.5 hours/day, sound level = 97 dB
Increase in sound intensity = final sound level - original sound level
Increase in sound intensity = 116 dB - 97 dB = 19 dB
Factor increase = 19/97 = 0.196 = 19.6%
Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
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The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5 m. If the box has a mass of 1 kg, what is the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s?
The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
Acceleration of the box
The acceleration of the box is calculated as follows;
vf² = vi² + 2as
a = (vf² - vi²)/2s
a = (11.5² - 13²) / (2 x 8.5)
a = -2.16 m/s²
Time of motion of the boxThe time taken for the box to travel is calculated as follows;
a = (vf - vi)/t
t = (vf - vi) / a
t = (11.5 - 13) / (-2.16)
t = 0.69 s
Average power supplied by the frictionP = Fv
P = (ma)(vf - vi)
P = (1 x -2.16) x (11.5 - 13)
P = 3.24 W
Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
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You are designing a delivery ramp for crates containing exercise equipment. The 1390 N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0∘. The ramp exerts a 515 N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp.
The maximum force constant of the spring Kmax is 2337.9 N/m.
What is force constant of a spring?The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.
Force constant is denoted by K and its unit is N/m.
Force = K * xWhere;
K = spring constant
x = displacement
The work done by the spring is given below as follows:
Work done = Fx/2
Kinetic Energy = mv²/2
Force on an inclined plane = mgsinθ
Total force, F = mgsinθ + frictional force
F = 1390 * sin 22° + 515
F = 1035.7 N
Work done = change in KE
Fx/2 = mv²/2
Fx = mv²
m = 1390/9.81 = 141.692
Solving for x;
x = mv²/F
x = 141.692 * 1.8²/1035.7
x = 0.443 m
The maximum force constant of the spring Kmax = 1035.7/0.443
Kmax = 2337.9 N/m
In conclusion, the maximum force constant of the spring is the ratio of the total force and displacement.
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Note that the complete question is given below:
You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria
Standing 38.6 m away from a rock wall, you yell. How much time in seconds will it take you to hear your echo to two significant digits? Make sure to account for the travel from you to the wall and from the wall back to you.
It will take you 0.23 seconds to hear your echo
What is Echo ?Echo can be simply defined as the reflection of sound wave.
Given that you are 38.6 m away from a rock wall, you yell. To know how much time in seconds it will take you to hear your echo to two significant digits, You must make sure to account for the travel from you to the wall and from the wall back to you.
The speed of sound V = 340 m/sThe distance D to and fro = 2 x 38.6 = 77.2 mThe time taken T = ?Speed V is the distance per time
V = D / T
Substitute all the parameters into the formula
340 = 77.2 / T
Make T the subject of formula
T = 77.2 / 340
T = 0.2270
T = 0.23 s to two significant digits
Therefore, it will take you 0.23 seconds to hear your echo.
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The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS
a.
i. PU is greater than PQ.ii. VU is Greater than 2VTb.
i. PR is Equal to PU.ii. VR is Equal to VSc.
i. VQ is Equal to VUii. PR is Greater than PS.What is pressure?Pressure is the force per unit area on a surface.
What is speed?Speed is the distance moved per unit time.
Pressure
Since pressure, P = hρg where
h = depth, ρ = density of liquid and g = acceleration due to gravity.Since ρ and g are constant
P ∝ h
So, we see that pressure is directly proportional to depth.
a. i. Pressure between R and USince U is lower than Q, Pressure at U is greater than pressure at Q.
So,PU is greater than PQ.
ii. Speed between U and TUsing the continuity equation
VUAU = VTAT where
VU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmVUT= speed at T, AT = cross-sectional area at T = π(dT)² where dT = diameter at T = 2.0 mmSo, VUAU = VTAT
VUπ(dU)² = VTπ(dT)²
VU = VT(dT)²/(dU)²
VU = VT(2.0)²/(1.0)²
VU = VT(4)
VU = 4VT
Since VU = 4VT,VU is Greater than 2VT
b i. Pressure between R and USince R is at the same depth as U, Pressure at R is equal to pressure at U.
So,PR is Equal to PU.
ii. Speed between R and SUsing the continuity equation
VRAR = VSAS where
VR = speed at R, AR = cross-sectional area at R = π(dR)² where dR = diameter at R = 2.0 mmVS= speed at S, AS = cross-sectional area at S = π(dS)² where dS = diameter at S = 2.0 mmSo, VRAR = VSAS
VRπ(dR)² = VSπ(dS)²
VR = VS(dS)²/(dS)²
VR = VS(2.0)²/(2.0)²
VR = VS(1)
VR = VS
Since VR = VS,VR is Equal to VS
c. i. Speed between Q and UUsing the continuity equation
VQAQ = VUAU where
VQ = speed at Q, AQ = cross-sectional area at Q = π(dQ)² where dQ = diameter at Q = 1.0 mmVU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmSo, VQAQ = VUAU
VQπ(dQ)² = VUπ(dU)²
VQ = VU(dU)²/(dQ)²
VQ = VU(1.0)²/(1.0)²
VQ = VU(1)
VQ = VU
Since VQ = VU, VQ is Equal to VU
Ii. Pressure between R and SSince R is lower than S, Pressure at R is greater than pressure at S.
So,PR is Greater than PS.
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a ball rolls off a platform that is 12 m above ground. The ball's horizontal velocity as it leaves the platform is 5 m/s. How much time it takes for the ball to hit the ground
The time it takes for the ball to hit the ground from the platform is determined as 1.56 seconds.
Time of motion of the ball
The time of motion of the ball is calculated as follows;
h = vt + ¹/₂gt²
where;
v is initial vertical velocity = 0t is time of motionh is height of fall of the ballh = 0 + ¹/₂gt²
h = ¹/₂gt²
t = √(2h/g)
t = √(2 x 12 / 9.8)
t = 1.56 seconds
Thus, the time it takes for the ball to hit the ground from the platform is determined as 1.56 seconds.
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An object is thrown upward with initial velocity of 30m/s at angle of 30 degree to the horizontal. calculate the components of the initial velocity.
Answer:
Vo (vertical) = Vo sin 30 = 30 m/s / 2 = 15 m/s
Vo (horizontal) = Vo cos 30 = .866 30 m/s = 26 m/s
Only using dimensional analysis, answer the following question:
Convert 65 miles per hour to kilometers per second. Show your work and explain how you got it please!
Brainliest goes to most detailed response
Answer:
[tex]2.9055\times 10^-2 km/s \approx 0.03 km/s[/tex]
Explanation:
One (statute) mile is defined as [tex]1\ 609.344 m = 1.609\ 334 km[/tex] or, cutting the last significant digits, [tex]1.6 km[/tex]. An annoying European would drop some kind of remark here like "but what about nautical miles?", but let's assume it's a statute mile.
One hour is 3600 seconds (60 minutes in a hour, times 60 seconds in a minute), no complains across the pond here.
Now we're ready to crunch numbers.
[tex]65\frac{mi}h = 65\frac{1.6km}{3600s} \approx 0.03 \frac {km}s[/tex]
You can use the exact definition of mile and you will get the exact value of [tex]0.029055 \frac{km}s = 2.9055\times10^-^2 km/s[/tex] with an error of about a meter.
Answer:
See below
Explanation:
Since you want detailed dimensional analysis here is a lot of details:
mile / hr / (60 min / hr * 60 sec / min) * 1760 yds/mile * 36 in/yd * 1m /39.37 in * 1 km / 1000 m
After cancelling everything out (dimensional analysis) , you are left with:
= mile/ hr * 1/3600 * 1760 * 36 /39.37 * 1/1000
= mile /hr * .00044704089 km -hr/ sec-mile = km/s
the underlined portion is the conversion factor for miles/hr to km/s
65 * .00044704089 = .029057 km / sec
A thin-walled hollow sphere has a radius 4cm from the center
of the sphere, the eletric field points radially inward and
has a magnitude of 1.5 × 10^4NC^-1 How much charge is on the surface
A thin-walled hollow sphere has a radius of 4 cm from the center
of the sphere, the electric field points radially inward and
has a magnitude of 1.5 × 10⁴ N/C then the charge on the surface would be 2.6690×10⁻⁹ C
What is an electric charge?Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.
The electric field inside a spherical shell is given by the formula
E = q/4πεr²
where q is the charge on the surface
r is the radius of the sphere
ε is the electrical permeability
By substituting the respective values in the given formula
1.5 × 10⁴ = q/4π(8.85✕ 10⁻⁻¹²)(.04²)
q = 2.6690×10⁻⁹ C
Thus, the charge on the surface would be2.6690×10⁻⁹ C
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A -4.00 nC point charge is at the origin, and a second -6.00 nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x -axis at x = 0.200 m'
The net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.
Force on electron due to charge 1
The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;
F = kq₀q₁/r²
where;
k is coulomb's constantq0 is charge at the originq1 is the charge at 0.2 m (electron)r is the distance between the chargesF(01) = (9 x 10⁹ x 4 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.2²)
F(01) = 1.44 x 10⁻¹⁶ N
Force on electron due to charge 2The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;
F = kq1q2/r²
where;
k is coulomb's constantq2 is charge at the 0.8 mq1 is the charge at 0.2 m (electron)r is the distance between the charges = 0.8 m - 0.2 m = 0.6 mF(12) = (9 x 10⁹ x 6 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.6²)
F(12) = 2.4 x 10⁻¹⁷ N
Net force on the electronF(net) = 2.4 x 10⁻¹⁷ N + 1.44 x 10⁻¹⁶ N
F(net) = 1.68 x 10⁻¹⁶ N
Thus, the net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.
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As the resistance increases, what happens to the slope on the current vs voltage graph?
A. It gets steeper
B. It gets flatter
C. It keeps a constant slope
D. Not enough info
B. As the resistance increases, the slope on the current vs voltage graph gets flatter.
What is resistance?Resistance is the opposition to the follow of current in electrical material.
The mathematical relationship between resistance, current and voltage is given by Ohm's law.
R = V/I
where;
R is resistance, measured in ohmsV is voltage, measured in voltsI is current, measured in ampereFrom the equation above, increase in resistance means an increase in voltage or decrease in current.
In Current vs voltage graph, the following will occur;
The current decreases - the vertical axis decreasesThe voltage increases - the horizontal axis becomes largerThus, as the resistance increases, the slope on the current vs voltage graph gets flatter.
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define universal vibrations ?
Answer:
The second universal law, the law of vibration, posits that everything (every atom, object, and living thing) is in constant motion, vibrating at a specific frequency.
Explanation:
Please mark brainly
The second universal law defines this.
The second universal law, also known as the Law of Vibration, The Law of Vibration states that everything in the universe is in a constant state of movement. We refer to these movements as vibration, and the speed or rate at which something vibrates is called its frequency. The only difference between one object and another is its vibration rate.
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A uniform plank of length 2.00 m and mass 30.23 kg is supported by three ropes. A 700 N person is a distance, of 0.55 m from the left end. A) find the magnitude of the tension in the vertical rope on the left end. Give your answers in newtons. B) find the magnitude of the tension in the rope in the right end. Give your answers in newtons C) find the magnitude of the tension in the horizontal rope on the left end. Give your answers in newtons
a. The magnitude of the tension, T₂ in the vertical rope on the left end is T₂ = 655.62 N
b. The magnitude of the tension in the rope in the right end is T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is T₃ = 406.3 N
What is tension force?Tension force refers to a pulling force that is exerted by a string or cable about an axis.
a. The magnitude of the tension, T₂ in the vertical rope on the left end is given as follows:
Taking moment about the vertical axis
T₂ = 30.23 * 9.81 + 700 - T₁ * Sin40°
Solving for T₁ by taking the left end as the pivot;
T₁ Sin 40° * 2.00 = 700 * 0.55 + (30.23 * 9.81) * 1.0
T₁ * 1.285 = 681.5563
T₁ = 530.4 N
Therefore;
T₂ = 30.23 * 9.81 + 700 - 530.4 * Sin 40°
T₂ = 655.62 N
b. From calculation, the magnitude of the tension in the rope in the right end is T₁.
T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is determined thus:
Taking moments about the left end in the horizontal direction;
T₃ = T₁ * cos 40°
T₃ = 530.4 N * cos 40°
T₃ = 406.3 N
In conclusion, the tension at the rope in the various ends is determined by taking moments about the left end.
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PLEASE HELP!!!!
Compared to driving at 20 mph, about how much longer will it take for you to stop at 60 mph?
about nine times as far
about six times as far
about twice as far
Answer:
The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)
Explanation:
Let [tex]u[/tex] denote the initial velocity of the vehicle ([tex]20\; \text{mph}[/tex] or [tex]60\; \text{mph}[/tex]) and let [tex]v[/tex] denote the velocity of the vehicle after braking ([tex]0\; \text{mph}[/tex]). Let [tex]x[/tex] denote the braking distance.
Assume that the acceleration during braking are both constantly [tex]a[/tex] in both scenarios. The SUVAT equations would apply. In particular:
[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}[/tex].
Since [tex]v = 0[/tex] (the vehicle has completely stopped), the equation becomes [tex]x = (-u^{2}) / (2\, a)[/tex].
Assuming that [tex]a[/tex] (braking acceleration) stays the same, the braking distance [tex]x[/tex] would be proportional to [tex]u^{2}[/tex], the square of the initial velocity.
Hence, increasing the initial speed from [tex]20\; \text{mph}[/tex] to [tex]60\; \text{mph}[/tex] would increase the braking distance by a factor of [tex]3^{2} = 9[/tex].
Answer:
9 times
Explanation:
The resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20 a, what is the voltage across the wire?
The voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V
What does the Resistance of a wire depend on ?The resistance of a wire is the opposition to the flow of current. It depends on the following;
TemperatureLength of the wireCross sectional areaResistivity of the wireGiven that the resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20A
The given parameters are;
Resistivity ρ = 1.72 × 10-8 ωm Length L = 1.0 mCross sectional area A = 2.0 × [tex]10^{-6}[/tex] m²Current I = 0.2 AResistance R = ?Voltage V = ?The formula to use to get R will be
R = ρL / A
Substitute all the necessary parameters into the formula
R = 1.72 x [tex]10^{-8}[/tex] x 1 / 2 x [tex]10^{-6}[/tex]
R = 8.6 x [tex]10^{-3}[/tex] Ω
From Ohm's law, V = IR
Substitute all the necessary parameters into the formula
V = 0.2 x 8.6 × [tex]10^{-3}[/tex]
V = 1.72 x [tex]10^{-3}[/tex] V
Therefore, the voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V
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Solve this question please A.S.A.P ⁉️
No spamming ❗
Answer:
Here given is :-Velocity (u) = 10m/sGravity (g) = 10 m/s²Height (h) = ?Explanation:
v² - u² = 2gh0² - 10² = 2 × (-10) × h-100 = -20 h[tex]h = \frac{ - 100}{ - 20} = 5m[/tex]V = u + gt0 = 10 - 10t10t = 10[tex]t = \frac{10}{10} = 1 \: sec[/tex]Total time = 2 secAnswer:
See below
Explanation:
vf = vo + at when it reaches top vf = 0 vo = 10m/s a = -9.81 m/s^2
0 = 10 + (-9.81) t
t = 1.02 seconds to reach max
time up = time down so roundtrip = 2.04 seconds
Two uncharged spheres are separated by 1.70 m. If 2.40 ✕ 10¹² electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
_______N
**Hint** Find the net charge on each sphere and substitute values into Coulomb's law.
The magnitude of the Coulomb force (in N) on one of the spheres, given the data is 4.59×10⁻⁴ N
How to determine the charge on each spheresSphere 1 losses 2.40×10¹² electrons
But
1 electron = 1.6x10¯¹⁹ C
Thus,
Charge on sphere 1 = +1.6x10¯¹⁹ × 2.40×10¹² = +3.84×10¯⁷ C
Sphere 2 gains 2.40×10¹² electrons
But
1 electron = 1.6x10¯¹⁹ C
Thus,
Charge on sphere 2 = -1.6x10¯¹⁹ × 2.40×10¹² = -3.84×10¯⁷ C
How to determine the coulomb forceCharge on sphere 1 (q₁) = +3.84×10¯⁷ CCharge on sphere 2 (q₂) = 3.60 mC = -3.84×10¯⁷ CElectric constant (K) = 9×10⁹ Nm²/C²Distance apart (r) = 1.7 mForce (F) =?
Using the Coulomb's law equation, the force can be obtained as illustrated below:
F = Kq₁q₂ / r²
F = (9×10⁹ × 3.84×10¯⁷ × 3.84×10¯⁷) / (1.7)²
F = 4.59×10⁻⁴ N
Thus, the magnitude of the Coulomb's force is 4.59×10⁻⁴ N
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According to current evidence, what happens when gays or lesbians rear a child?
a. the child is more likely than average to develop depression or anxiety disorders.
O b. the child develops about the same as other children.
O c. the child is more likely than average to become gay or lesbian.
O d. the child does poorly in school.
Answer: The answer is A.
Explanation:
Just because u have gay parents doesnt mean theyre bad lol XD
An Airbus A380-800 passenger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m^2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? (The average density of the air around the airplane is ρair = 1.21 kg/m^3.)
The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.
We need to be aware of the Bernoulli principle in order to determine the solution.
How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:[tex]P+\frac{1}{2}dv^2+ dgh = constant.[/tex] We substitute d for to represent density.
We've done that,[tex]V_1=259m/s\\V_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]
We compare the governing idea for the wing's bottom and upper surfaces to:[tex]P_1+\frac{1}{2}dV_1^2+dgh=P_2+ \frac{1}{2}dV_2^2+dgh\\P_1-P_2=\frac{1}{2}d(V_2^2-V_1^2)\\\frac{F}{A}= \frac{1}{2}d(V_2^2-V_1^2)\\[/tex]
Consequently, using the aforementioned equation, the airplane's mass will be,[tex]m=\frac{\frac{1}{2}d(V_2^2-V_1^2)A\\}{g} \\m=387*10^3kg.[/tex]
Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.
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A 92-kg fullback is running at 3.6 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west.
a)Calculate the original momentum of the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the momentum is to the east, and negative value if the direction of the momentum is to the west.
b)Calculate the impulse exerted on the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
c)Calculate the impulse exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
d)Calculate the average force exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the force is to the east, and negative value if the direction of the force is to the west.
a) The original momentum of the fullback is 331.2 kgm/s.
b) The impulse exerted on the fullback is - 331.2 kgm/s
c) The impulse exerted on the tackler is 331.2 kgm/s
d) The average force exerted on the tackler is 389.64 N
Given:Mass of the fullback, m = 92 kg
Initial velocity of the fullback , u = 3.6 m/s
Time of the motion of the fullback , t = 0.85s
The original momentum of the fullback ;
[tex]P_{i} = mv\\P_{i} = (92) (3.6)\\P_{i} = 331.2 kg m/s[/tex]
The impulse exerted on the full-back;
J = ΔP = [tex]m v_f - m v_i[/tex]
J = [tex]m ( v_f - v_i)[/tex]
J = 92 ( 0- 3.6)
J = - 331.2 kgm/s
The impulse exerted on the tackler;
[tex]J _1 = - J _2\\J_2 = - J_1\\J_2 = -( - 331.2)\\J_2 = 331.2 kgm/s[/tex]
The average force exerted on the tackler;
F= Δmv/t
[tex]F =\frac{331.2}{0.85} \\F = 389.64 N[/tex]
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Please help with this question, step by step
Thank you!
From the calculation, the force constant is 192 N. Also, friction would decrease the extension.
What is the force constant?We know that the force constant can be obtained by the use of the relation;
F = Ke
F = applied force
K = force constant
e = extension
We know from Hooks law that the force applied is directly proportional to the extension.
We can write;
F = mgcosθ
F = 43 Kg * 9.8 m/s^2 * sin31°
F = 217 N
K = 217 N/1.13 m
K = 192 N/m
If there is friction between the incline and the crate, it will stretch less because some work will be lost due to friction causing only some fraction of the elastic potential energy to be converted to kinetic energy.
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is a pot plant a closed system?
Answer:
No because a pot plant system must interact with its surroundings (air, water, etc) and these are not closed systems.
Only if the pot plant were completely enclosed, say in a large glass jar would the system be closed.
A ball of mass 0.200 kg has a velocity of 150m/s; a ball of mass 0.300 kg has a velocity of - 0.4m/s. They meet in a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of mass before and after the collision.
(a) The velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.
(b) The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.
Velocity of the balls after collision
Apply the principle of conservation of linear momentum to determine the velocity of the balls;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.2(150) + (0.3)(-0.4) = 0.2v₁ + 0.3v₂
29.88 = 0.2v₁ + 0.3v₂
Apply one directional linear velocity
u₁ + v₁ = u₂ + v₂
v₁ = u₂ + v₂ - u₁
v₁ = -0.4 + v₂ - 150
v₁ = v₂ - 150.4
Substitute the value of v₁ into the first equation;
29.88 = 0.2(v₂ - 150.4) + 0.3v₂
29.88 = 0.2v₂ - 30.08 + 0.3v₂
59.96 = 0.5v₂
v₂ = 59.96/0.5
v₂ = 119.92 m/s
v₁ = 119.92 - 150.4
v₁ = -30.48 m/s
Velocity of their center mass before collisionV(cm) = (0.2 x 150 + 0.3 x 0.4) / (0.2 + 0.3)
V(cm) = 60.24 m/s
Velocity of their center mass after collisionV(cm) = (0.2 x 30.48 + 0.3 x 119.92) / (0.2 + 0.3)
V(cm) = 84.14 m/s
Thus, the velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.
The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.
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Consider the f(x) = cos(x) + D function shown in the figure in blue color. What is the value of parameter D?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The value of parameter D is -3.
What is angular frequency of a wave?Angular frequency is the angular displacement of any element of the wave per unit time.
f(x) = Acos(x) + D
where;
A is amplitude of the waveB is phase difference of the waveWhat is amplitude of a wave?The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
From the blue colored graph, y = -2 at x = 0
-2 = cos(0) + D
-2 = 1 + D
-2 - 1 = D
-3 = D
Thus, the value of parameter D is -3.
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1. The diagram shows a satellite traveling in uniform circular motion around the Earth.
(a) Give the relation between radius of the orbit and the velocity of the satellite.
(b ) The satellite is kept in orbit by a force. On the diagram draw an arrow to show the
direction of this force.
Answer:
M V R = constant angular momentum is constant because no forces act in the direction of V
Since M (mass) = constant
V R = constant
The force is directed along the gravitational force vector (towards the center of rotation)
