the potential energy of two particles when they are brought together to a distance of 1.2x10^-15 m, the approximate size of a nucleus.The equation for potential energy between two charged particles is:U = (k * q1 * q2) / rwhere U is the potential energy, k is Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them. The distance between the two particles is 1.2 × 10^-15 m.
The size of the nucleus is roughly 1.2 × 10^-15 m, so the distance between the particles is the same as the size of the nucleus.Now, let's assume that the particles are protons, which have a charge of +1.6 × 10^-19 C each. We'll use Coulomb's constant, k = 9 × 10^9 N·m²/C², which gives us:U = (9 × 10^9 N·m²/C²) × ((1.6 × 10^-19 C)²) / (1.2 × 10^-15 m)U = 2.3 × 10^-13 JTherefore, the potential energy of the two particles when they are brought together to a distance of 1.2 × 10^-15 m is 2.3 × 10^-13 J.
Potential energy is the energy that an object possesses because of its position in a force field or that a system possesses because of the configuration of its parts. The potential energy of two charged particles is the energy that the two particles possess due to their position with respect to each other. The potential energy of two particles is directly proportional to the product of their charges and inversely proportional to the distance between them. When the two particles are brought closer together, their potential energy increases. When they are moved farther apart, their potential energy decreases.
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A cargo plane is flying horizontally at an
altitude of 14.7 km with a constant speed of
994 km/hr when a large crate slides out of
the rear of the plane. A conceptual sketch is
shown.
994 km/hr
14.7 km
x
How long does it take the crate to hit
the ground? The acceleration of gravity is
9.81 m/s
2
.
1. 54.7443
2. 47.3562
3. 51.0841
4. 49.4619
5. 48.8397
6. 52.2676
7. 45.3776
8. 53.9944
9. 53.425
10. 50.0764
Answer in units of s
Answer:
1. 54.74433 s
Explanation:
h = Height of plane = 14.7 km
[tex]u_x[/tex] = Velocity of crate in x direction = 994 km/h = [tex]276.11\ \text{m/s}[/tex]
a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
[tex]h=u_yt+\dfrac{1}{2}at^2\\\Rightarrow t=\sqrt{\dfrac{2h}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 14700}{9.81}}\\\Rightarrow t=54.74433\ \text{s}[/tex]
Time the crate takes to hit the ground is 54.74433 s
14. a 60.0 w lamp is placed in series with a resistor and a 120.0 v source. if the voltage across the lamp is 25 v, what is the resistance of the resistor?
The resistance of the resistor in the series circuit can be calculated using Ohm's Law. The resistance of the resistor is approximately 200 Ω.
Given values:
Power of the lamp (P) = 60.0 W
Voltage across the lamp (Vlamp) = 25 V
Voltage of the source (Vs) = 120.0 V
Using the formula for power: Power = Voltage × Current
We can rearrange the formula to find the current (I) flowing through the circuit: I = Power / Voltage
Substituting the values, we find the current flowing through the circuit:
Current (I) = 60.0 W / 120.0 V
Current (I) = 0.5 A
Since the lamp and the resistor are in series, the current passing through both components is the same.
Voltage across the resistor (Vresistor) can be calculated using Ohm's Law: Vresistor = I × R
Rearranging the formula, we find the resistance (R) of the resistor: R = Vresistor / I
Substituting the values, we can calculate the resistance of the resistor:
Resistance (R) = 25 V / 0.5 A
Resistance (R) = 50 Ω
Therefore, the resistance of the resistor in the series circuit is approximately 200 Ω.
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Explain how each of the following changes as you travel across the spectrum: wavelength, frequency, and energy
The frequency and energy increases while the wawelength decreases on moving from left to right.
Since the energy of the wave is directly proportion to the frequency and indirectly proportionl to the wavelength.
On moving from left to right (Gamma to radio waves), the wavelenth in creases.On moving from left to right (Gamma to radio waves), frequency decreases.On moving from left to right (Gamma to radio waves), energy decreases.To know more about energy of the wave,
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A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south. What is the charge of object A? a. 1.11 Times 10^-9 C b. -1.11 Times 10^-9 C c. 2.78 Times 10^-10 C d. -2.78 Times 10^-10 C e. 5.75 Times 10^12 C f. -5.75 Times 10^12 C
The electric field created by a point charge is given by the equation E = k * (Q / [tex]r^{2}[/tex]).The charge of object A is -1.11 × [tex]10^{-9}[/tex] C (option b).
In this case, the electric field at point P is 40.0 N/C directed to the south. Since the field is directed towards the south, the charge of object A must be negative. Plugging the given values into the equation, we have:
40.0 N/C = (9 × [tex]10^{9}[/tex] N [tex]m^2/C^2[/tex]) * (Q / (0.250 [tex]m)^2)[/tex]
Simplifying the equation, we can solve for Q:
Q = (40.0 N/C) * (0.250 [tex]m)^2[/tex] / (9 × 10^9 N [tex]m^2/C^2)[/tex]
Calculating the expression, we find Q ≈ -1.11 × [tex]10^{-9}[/tex] C. Therefore, the charge of object A is approximately -1.11 × [tex]10^{-9}[/tex] C, which corresponds to option b.
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The ____ will convert the electrical energy to some other form of energy.
a load
b battery
c wire
d conductor
Answer:
I believe it d I'm not completely sure but yeah
A rock climber, of total mass 62kg, holds herself in horizontal equilibrium against a vertical cliff. She pulls
on a rope that is fixed at the top of the cliff and presses her feet against the cliff.
(a) Calculate the total weight of the climber.
(b) State the two conditions needed for equilibrium.
(c) The climber’s centre of mass is 0.90m from the cliff.
(i) Calculate the moment about her feet due to her weight
(ii) The line of the rope meets the horizontal line through her centre of mass at distance of 1.2m from the
cliff, as shown in the figure. The rope is at an angle of 60° to the horizontal. Determine the tension in the
rope. (Take g = 10ms-2
)
Answer:
(a) The total weight of the climber is equal to her mass multiplied by the acceleration due to gravity. Therefore, W = mg = 62 kg x 10 m/s^2 = 620 N.
(b) The two conditions needed for equilibrium are that the net force acting on the climber is zero and the net torque acting on the climber is zero.
(c)(i) The moment about her feet due to her weight is equal to the weight of the climber multiplied by the distance between her feet and the cliff. Therefore, M = W x d = 620 N x 0.9 m = 558 Nm.
(ii) To determine the tension in the rope, we need to resolve the forces acting on the climber in the horizontal and vertical directions. In the horizontal direction, the tension in the rope is balanced by the force of friction between the climber's feet and the cliff. Therefore, T = F.
In the vertical direction, the climber's weight is balanced by the normal force of the cliff and the tension in the rope. Therefore, N + Tcos(60) = W.
Since the climber is in equilibrium, the net torque acting on her must be zero. Therefore, the torque due to the tension in the rope must be equal and opposite to the torque due to the climber's weight. Therefore, Tsin(60) x 1.2 = M.
Substituting the values we have, we get:
N + Tcos(60) = W
Tsin(60) x 1.2 = M
Solving for T, we get:
N = W - Tcos(60) = 620 N - T(0.5)
Substituting this into the second equation, we get:
Tsin(60) x 1.2 = M
Tsin(60) = M / 1.2 = 558 Nm / 1.2 m = 465 N
Substituting this value of T into the first equation, we get:
N = 620 N - T(0.5) = 620 N - 465 N(0.5) = 388 N
Therefore, the tension in the rope is 465 N and the normal force of the cliff on the climber is 388N
if you gently push a ball off of a table top, 1.3 m above the floor. how long (in seconds) does it take the ball to reach the floor?
It will takes approximately 0.165 seconds for the ball to reach the floor after it has been gently pushed off the table top from a height of 1.3 m above the floor.
When an object is dropped from a height, it falls to the ground due to the force of gravity.
When an object is released from a height of 1.3 m above the ground, the time it takes to reach the ground is calculated using the formula:
h = 0.5gt² where h is the height of the object, g is the acceleration due to gravity and t is the time taken for the object to reach the ground.
Rearranging this formula to solve for t, we have: [tex]t = \sqrt{(2h/g)}[/tex]
Substituting the values we have: [tex]h = 1.3 mg = 9.81 m/s^{2} t = \sqrt{(2 \times 1.3 / 9.81)t} = \sqrt{(0.265 / 9.81)t} = \sqrt{0.027t } = 0.165 seconds[/tex]
Therefore, it takes approximately 0.165 seconds for the ball to reach the floor after it has been gently pushed off the table top from a height of 1.3 m above the floor.
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a solid plastic cylinder of radius 2.33 cm and length 6.30 cm. find the charge of this cylinder, in nc (nanocoulombs), given that its has a uniform surface charge density of 13.8 nc/m2 everywhere on its surface.
The charge of this cylinder is calculated as 10.2 nC. The formula for the charge of a cylinder with uniform surface charge density is ; Q = (charge density) * (surface area) * (length).
Charge density = σ,
Surface area = 2πrh + 2πr²,
Length = l
Given : Radius, r = 2.33 cm, Length, l = 6.30 cm,
Charge density, σ = 13.8 n c/m²,
Surface area of cylinder = 2πrh + 2πr² where; h = l (length)
Substitute the value of radius and height in the above formula, we get;
Surface area of cylinder = 2πr(l + r)
Surface area of cylinder = 2π(2.33 cm)(6.30 cm + 2.33 cm)
= 119.38 cm²
Charge density, σ = 13.8 nc/m²
= 1.38 × 10⁻⁵ C/cm²
Now, put the value of σ, surface area, and length in the formula to calculate the charge of cylinder as;
Q = σ * surface area * length
Q = (1.38 × 10⁻⁵ C/cm²) * (119.38 cm²) * (6.30 cm)
Q = 1.02 × 10⁻² C = 10.2 nC
Therefore, the charge of this cylinder is 10.2 nC (nanocoulombs).
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Irreversible damage to the liver due to long term alcohol abuse is called:
A.Cirrhosis
B. Cancer
C. Liver failure
D. jaundice
Answer:
A, Cirrhosis
Explanation:
What is the density of mercury if 67.67g fills a 2mL graduated cylinder? Include the unit
for full credit (remember the unit is mass unit /volume unit).
Answer:
[tex]Density = 33835kg/m^3[/tex]
Explanation:
Given
[tex]Mass = 67.67g[/tex]
[tex]Volume = 2mL[/tex]
Required
Determine the density
[tex]Density = \frac{Mass}{Volume}[/tex]
First, we need to convert mass to kg
[tex]Mass = 67.67g[/tex]
[tex]Mass = \frac{67.67kg}{1000}[/tex]
[tex]Mass = 0.06767kg[/tex]
Next, we need to convert volume to m^3
[tex]Volume = 2mL[/tex]
[tex]Volume = 2 * 10^{-6}m^3[/tex]
Density is then calculated as follows:
[tex]Density = \frac{0.06767}{2 * 10^{-6}}[/tex]
[tex]Density = \frac{0.06767 * 10^6}{2}[/tex]
[tex]Density = \frac{67670}{2}[/tex]
[tex]Density = 33835kg/m^3[/tex]
a magnet has stronger force of attraction at the poles than in the middle why?
Answer:
The molecular magnets are arranged in an open chain so that the north pole or the south pole of the molecular magnets lie in the same direction which gives strong force at the poles whereas two opposite poles are arranged at the middle and the force cancel each other. So, poles have more force than the middle portion.
The magnet has stronger force of attraction at the poles than in middle, because, the at the poles they are attracted to the opposite poles. Whereas at the middle, the magnetic dipoles are cancelled each other and have no more force of attraction.
What is a magnet?A magnet is a material made of metals which are of ferromagnetic nature. Thus, produce permanent magnetic dipoles. The magnetic field lines produced from a permanent magnet is strong at can fluctuate with the presence of an electric field.
A bar magnet have two poles namely a south pole and north pole. Like two the electric charges, where two like charges repel each other and unlike charges attracts, two like poles of a magnet repels and unlike poles attracts each other.
Therefore, at the two poles the magnetic field lines towards the opposite poles will be more stronger whereas at the middles the field lines merges and are oriented randomly and have no strong magnetic effect.
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a photographer focuses his camera on his subject. the subject then moves closer to the camera. to refocus, should the lens be moved closer to or farther from the detector? explain.
A photographer focuses his camera on his subject. The subject then moves closer to the camera. To refocus, the lens should be moved further from the detector.
When the subject is moved closer to the camera, the camera's lens should be moved away from the detector. As the subject moves closer to the camera, the distance between the lens and the detector decreases, causing the camera to lose focus.
Therefore, in order to refocus the camera and capture a clear image of the subject, the lens must be moved further away from the detector, increasing the distance between the lens and the detector and reestablishing the proper focal length.
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a small container of water is placed on a carousel inside a microwave oven, at a radius of 11.5 cm from the center. the turntable rotates steadily, turning through one revolution each 6.50 s. what angle does the water surface make with the horizontal?
Angle made by the water surface with the horizontal is calculated as 6.28 radians. Angle made by the water surface with the horizontal is : Angular speed of the container x Time taken to complete one revolution.
Circumference of a circle = 2πr, where r is the radius of the circle and π is the mathematical constant pi (approximately 3.14).In this case, the radius of the circle is 11.5 cm.
Therefore, the circumference of the circle is:
Circumference = 2πr = 2 x 3.14 x 11.5 cm
= 72.38 cm
In one revolution, the container moves through an angle of 360°, which is the same as 2π radians.
Therefore, the angle through which the container moves in one revolution is: Angle moved in one revolution = 2π radians
Since the turntable rotates steadily, turning through one revolution each 6.50 s, the angular speed of the container can be found by dividing the angle moved in one revolution by the time taken to complete one revolution.
Angular speed of the container = Angle moved in one revolution / Time taken to complete one revolution
= 2π radians / 6.50 s
= 0.968 radians/s
Angle made by the water surface with the horizontal= Angular speed of the container x Time taken to complete one revolution
= 0.968 radians/s x 6.50 s
= 6.28 radians
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What is the equivalent resistance (total resistance) of the series circuit shown?
Answer:
In a series circuit, the equivalent resistance is the algebraic sum of the resistances. The current through the circuit can be found from Ohm's law and is equal to the voltage divided by the equivalent resistance. The potential drop across each resistor can be found using Ohm's law.
Typical electrical wires in your house are generally made of copper (electron density of 8.47∗1024 electrons per cm3 ) and are usually either 14 gauge (diameter of 1.63 mm ), 12 gauge (diameter of 2.05 mm ), or 10gauge(2.59 mm) wires.
(A) If you have a 14 gauge wire that is carrying the maximum current of 20 , what would be the average drift speed of the electrons in the wire? Tries 1/8
(B) Usinq the averaqe speed you just calculated in Part (A), how much time would it take an electron to travel 8.84 m?
A. The average drift speed of the electrons in the wire is 1.19 x [tex]10^{-3}[/tex]cm/s.
B. The time it would take an electron to travel 8.84 m is 7.43 x [tex]10^{5}[/tex]sec
(A) To find the average drift speed of electrons in a wire, we can use the equation:
I = nAvq
Where:
I is the current in Amperes
n is the electron density in electrons per [tex]cm^{3}[/tex]
A is the cross-sectional area of the wire in [tex]cm^{2}[/tex]
v is the average drift velocity of electrons in cm/s
q is the charge of an electron, which is 1.6 x [tex]10^{-19}[/tex] Coulombs
First, we need to find the cross-sectional area of the wire. The formula for the area of a circle is:
A = π[tex]r^{2}[/tex]
Where:
A is the area of the circle
r is the radius of the circle
Given that the wire diameter is 1.63 mm, we can find the radius by dividing it by 2:
r = 1.63 mm / 2 = 0.815 mm = 0.0815 cm
Now, we can calculate the area:
A = [tex]π(0.0815 cm)^{2}[/tex] = 0.0209 [tex]cm^{2}[/tex]
Next, we can rearrange the equation to solve for v:
v = I / (nAq)
Given that the current is 20 A and the electron density is 8.47 x [tex]10^{24}[/tex]electrons per [tex]cm^{3}[/tex], we can substitute these values into the equation:
v = 20 A / (8.47 x [tex]10^{24}[/tex] electrons per cm^3 * 0.0209 cm^2 * 1.6 x [tex]10^{-19}[/tex] C)
Simplifying the expression:
v = 1.19 x [tex]10^{-3}[/tex] cm/s
Therefore, the average drift speed of electrons in the 14 gauge wire carrying a maximum current of 20 A is approximately 1.19 x [tex]10^{-3}[/tex] cm/s.
(B) To calculate the time it would take for an electron to travel a distance of 8.84 m, we can use the formula:
t = d / v
Where:
t is the time in seconds
d is the distance in meters
v is the average drift velocity of electrons in meters per second
Given that the distance is 8.84 m and the average drift velocity is 1.19 x 10^-3 cm/s, we need to convert the velocity to meters per second:
v = 1.19 x [tex]10^{-3}[/tex] cm/s * 0.01 m/cm = 1.19 x [tex]10^{-5}[/tex] m/s
Now, we can substitute the values into the formula:
t = 8.84 m / 1.19 x [tex]10^{-5}[/tex]m/s
Simplifying the expression:
t = 7.43 x [tex]10^{5}[/tex]s
Therefore, it would take approximately 7.43 x [tex]10^{5}[/tex] seconds for an electron to travel a distance of 8.84 m.
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Please help me solve this problem on my homework!
Answer:
B cause u need to find the total range
Explanation:
find the most probable speed for nitrogen molecules at 295 k. group of answer choices 320 m/s 419 m/s 450 m/s 519 m/s
The most probable speed for nitrogen molecules at 295K can be found using the Maxwell-Boltzmann distribution of speeds equation. The equation is as follows:f(v) = 4π(\frac{m}{2πkT})^{\frac{3}{2}}v^2e^{\frac{-mv^2}{2kT}}$$where,m = \text{mass of one molecule of nitrogen}T = \text{temperature in Kelvi k = \text{Boltzmann constant}
Now, substituting the values for nitrogen molecules, we get,$$m = 4.65 × 10^{-26}\text{ kg}$$$$T = 295 \text{ K}k = 1.38 × 10^{-23}\text{ J/K}Substituting these values in the equation, we get,f(v) = 4π(\frac{4.65 × 10^{-26}}{2π×1.38 × 10^{-23}×295})^{\frac{3}{2}}v^2e^{\frac{-4.65 × 10^{-26}v^2}{2×1.38 × 10^{-23}×295}}Simplifying the expression gives:f(v) = 4.351 × 10^{-25}v^2e^{\frac{-5.022 × 10^{-27}v^2}{1}}We need to find the most probable speed, which is where the distribution function is maximum. Therefore, we differentiate the above equation w.r.t. v and equate it to zero.f'(v) = 0\frac{d}{dv}[4.351 × 10^{-25}v^2e^{\frac{-5.022 × 10^{-27}v^2}{1}}] = 08.702 × 10^{-25}v e^{\frac{-5.022 × 10^{-27}v^2}{1}}- 2.575 × 10^{-51}v^3 e^{\frac{-5.022 × 10^{-27}v^2}{1}}= 0$$$$v = \sqrt{\frac{2kT}{m}} = \sqrt{\frac{2×1.38 × 10^{-23}×295}{4.65 × 10^{-26}}} = 517.3 \text{ m/s}$$Therefore, the most probable speed for nitrogen molecules at 295K is 517.3 m/s (approx). Hence the closest option available is 519 m/s.For such more question on temperature
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The most probable speed for nitrogen molecules at 295 K can be calculated using the formula: VMP = √(2kT/m) T is the temperature in Kelvin, and m is the mass of the molecule in kilograms.
For nitrogen (N2),
m = 28 g/mol or 0.028 kg/mol. Now, let's plug in the values:
vmp = [tex]√(2(1.38 × 10^-23 J/K)(295 K)/(0.028 kg/mol)[/tex]
)vmp = [tex]√(8.038 × 10^-21 J/kg)[/tex]
vmp =[tex]8.96 × 10^2 m/s[/tex]
Therefore, the most probable speed for nitrogen molecules at 295 K is approximately 896 m/s. the most probable speed for nitrogen molecules at 295 K" is 896 m/s.
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Hello:)! how to do 1(c)?
Answer:
It does not come to a complete stop immediately because of inertia.
Explanation:
Inertia is the reluctance of an object to change its state of motion or rest. In this case, it is the state of motion since the ship was moving initially. Inertia is directly related to mass. Thus, the heavier the mass, the larger the inertia. The ship has a large mass and therefore it has a large inertia. This causes the ship to slowly come to a stop instead of stopping immediately when the engines are switched off.
Question 6 (15 points)
According to Newton's second law, when the same force is applied to two objects of
different masses,
the object with greater mass will experience a great acceleration, and the object
with less mass will experience an even greater acceleration.
the object with greater mass will experience a smaller acceleration, and the
object with less mass will experience a greater acceleration.
the object with greater mass will experience a greater acceleration, and the
object with less mass will experience a smaller acceleration.
the object with greater mass will experience a small acceleration, and the object
with less mass will experience an even smaller acceleration.
Answer:
that would be newtons 3rd law
Explanation:
because its how it is
The Navy Pier Ferris Wheel in Chicago has a circumference that is 56% of the circumference of the first Ferris wheel built in 1893. a. What is the radius of the Navy Pier Ferris Wheel? To the nearest foot, the radius of the Navy Pier Ferris Wheel is 1$$70 feet. b. What was the radius of the first Ferris wheel? To the nearest foot, the radius of the first Ferris wheel was 2$$125 feet. c. The first Ferris wheel took 9 minutes to make a complete revolution. How fast was the wheel moving? To the nearest tenth, the wheel was moving at a speed of 3$$87.2 feet per min
Answer: a. 70 ft
b. 125 ft
c. 87.2 ft/sec
Explanation:
Here is the complete question:
If the navy pier ferris wheel in chicago has a circumference that is 56% of the circunference of the first ferris wheel built in 1893.
a. What is the radius of the navy pier wheel?
b. what was the radius of the first ferris wheel?
c. The first ferri wheel took nine minutes to make a complete revolution. how fast was the wheel moving?
C=439.6 ft.
a. What is the radius of the navy pier wheel?
Since circumference = 2πr
where π = 3.142
r = radius
Therefore, 2πr = 439.6
(2×3.142×r) = 439.6
6.284r = 439.6
Radius = 439.6/6.284
Radius = 69.9 = 70 feet
b. What was the radius of the first ferris wheel?
Since the Navy Pier Ferris Wheel in Chicago has a circumference that is 56% of the circumference of the first Ferris wheel. This can be mathematically written as:
56% of f = 439.6
0.56f = 439.6
f = 439.6/0.56
f = 785
The circumference of the first wheel is 785ft
Radius will now be:
2πr = 785
r = 785/2π
r = 785/(2×3.142)
r = 785/6.284
r = 124.9 = 125ft
c. The first ferri wheel took nine minutes to make a complete revolution. how fast was the wheel moving?
= 785/9
= 87.2 ft/sec
what is a radio active element
A radioactive element, also known as a radionuclide or radioisotope, is an element that exhibits radioactivity.
Radioactive elements are the elements that have an unstable nucleus and emit radiation when their nucleus undergoes a process called radioactive decay. Radioactive decay is a random process that occurs in some of the elements with unstable atomic nuclei.
During this process, the nucleus emits alpha, beta, or gamma particles, which transform the nucleus into a more stable state. As a result, these elements are radioactive and can release high-energy particles or electromagnetic waves in the form of radiation, which can be dangerous for living organisms. Radioactive elements are mainly classified into two types: natural radioactive elements and artificially produced radioactive elements.
Natural radioactive elements are those that occur in nature, and their nuclei naturally undergo radioactive decay. For example, Uranium, thorium, and radium are natural radioactive elements. The artificially produced radioactive elements are those that are synthesized through nuclear reactions in a laboratory. For example, technetium and plutonium are artificially produced radioactive elements.
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Find the length of time required for the total pressure in a system containing N2Os at an initial pressure of 0.110 atm to rise to 0.145 atm Express your answer using two significant figures. Exercise 15.116 The specific rate constant for the first-order decomposition of N205 (g) to NO2 (g) and Orlg) is 7.48 × 10-3 s-1 at a given temperature t= 32 s Submit My Answers Give Up Correct Part B Find the length of time required for the total pressure in a system containing N2Os at an initial pressure of 0.110 atm to rise to 0.220 atm Express your answer using two significant figures. t- 150 Submit My Answers Give Up Correct Part C Find the total pressure after 110 s of reaction. atm Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining: no points deducted
a) The length of time required for the total pressure to rise from 0.110 atm to 0.145 atm is approximately 35.2 seconds.
b) The length of time required for the total pressure to rise from 0.110 atm to 0.220 atm is approximately 92.9 seconds.
c) Pf ≈ 0.054 atm
The total pressure after 110 seconds of reaction is approximately 0.054 atm.
To solve this problem, we can use the integrated rate law for a first-order reaction:
ln(Pf/Pi) = -kt
Where:
Pf is the final pressure
Pi is the initial pressure
k is the rate constant
t is the time
Let's solve each part of the problem:
Part A:
Using the given information:
Pi = 0.110 atm
Pf = 0.145 atm
k = 7.48 × 10^-3 s^-1
ln(Pf/Pi) = -kt
ln(0.145/0.110) = -(7.48 × 10^-3 s^-1) * t
0.263 = -7.48 × 10^-3 t
t ≈ 35.2 s
Part B:
Using the given information:
Pi = 0.110 atm
Pf = 0.220 atm
k = 7.48 × 10^-3 s^-1
ln(Pf/Pi) = -kt
ln(0.220/0.110) = -(7.48 × 10^-3 s^-1) * t
0.693 = -7.48 × 10^-3 t
t ≈ 92.9 s
Part C:
Using the given information:
t = 110 s
k = 7.48 × 10^-3 s^-1
We can rearrange the integrated rate law to solve for the final pressure (Pf):
ln(Pf/Pi) = -kt
Pf/Pi = e^(-kt)
Pf = Pi * e^(-kt)
Pf = (0.110 atm) * e^(-(7.48 × 10^-3 s^-1) * (110 s))
Pf ≈ 0.054 atm
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6. Draw conclusions: Newton’s first law states that an object in motion will travel at a constant velocity unless acted upon by an unbalanced force. How do these experiments show this?
The position an object moving along x-axis is given by x=a+bt², where a=38m and b=29m/s². What is the velocity at t=10s and t=50s? What is the velocity between 20s and 28s?
The position an object moving along x-axis is given by x=a+bt², where a=38m and b=29m/s². The velocity at t = 10 s is 580 m/s, at t = 50 s is 2900 m/s, and between 20 s and 28 s is 1392 m/s.
To find the velocity of the object at different time intervals, we need to take the derivative of the position equation with respect to time.
Given the position equation x = a + bt², where a = 38 m and b = 29 m/s², let's calculate the velocity at different time points.
Taking the derivative of the position equation, we get:
v = dx/dt = 2bt.
Substituting the given values of a and b, we have:
v = 2(29 m/s²)t.
Now we can calculate the velocity at specific time points:
a) At t = 10 s:
v = 2(29 m/s²)(10 s) = 580 m/s.
b) At t = 50 s:
v = 2(29 m/s²)(50 s) = 2900 m/s.
c) Between t = 20 s and t = 28 s:
To find the velocity between these time intervals, we need to find the difference in position and divide it by the difference in time.
Position at t = 20 s:
x1 = a + b(20 s)² = 38 m + (29 m/s²)(400 s²) = 38 m + 11600 m = 11638 m.
Position at t = 28 s:
x2 = a + b(28 s)² = 38 m + (29 m/s²)(784 s²) = 38 m + 22736 m = 22774 m.
Velocity between 20 s and 28 s:
v = (x2 - x1) / (t2 - t1) = (22774 m - 11638 m) / (28 s - 20 s) = 11136 m / 8 s = 1392 m/s.
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Why ground roll cannot be separated from reflectors in F-K
domain?
In the F-K domain, ground roll cannot be separated from reflectors as it refers to the low-frequency noise or energy that is associated with seismic waves propagating through the near-surface layers of the Earth.
It typically appears as a coherent and continuous signal in the seismic data. In the F-K domain, which represents the data in terms of frequency and wavenumber, the ground roll and the reflectors share similar characteristics in terms of their frequency content and wavenumber distribution. As a result, it becomes challenging to separate the ground roll from the reflectors based solely on their F-K domain representation.
The F-K domain provides information about the frequency and spatial characteristics of the seismic data. However, it does not provide direct information about the physical properties or origins of the seismic energy. Both ground roll and reflectors can have similar frequency-wavenumber signatures, making it difficult to distinguish between them in this domain. Additional processing techniques and analysis, such as filtering or velocity analysis, are often required to separate the ground roll from the desired reflectors and enhance the interpretability of the seismic data.
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What is Newton's 2nd law of motion? *
Your answer
Answer:
Newton's second law describes the affect of net force and mass upon the acceleration of an object.
F = force
m = mass of an object
a = acceleration
Explanation:
f=m.a
what temperature would be required to double the volume of an ideal gas originally at stp if the pressure decreased by 25%
The temperature required to double the volume of an ideal gas originally at STP if the pressure decreased by 25% would be 546K. STP stands for Standard Temperature and Pressure. It is a standard set of conditions for measuring and comparing gases.
The standard temperature is 0°C or 273K, and the standard pressure is 1 atm or 101.325 kPa. STP is frequently utilized in scientific research and industry to guarantee that data can be compared. Charles' law is a gas law that states that the volume of a given mass of an ideal gas at constant pressure is directly proportional to its temperature in Kelvin. It is stated mathematically as V / T = constant. Where V is the volume of the gas, and T is its temperature in Kelvin. Charles' law may be used to calculate the volume of a gas when the temperature changes if the initial and final temperatures are known.
Charles' law may be written as V₁ / T₁ = V₂ / T₂, where V₁ and T₁ are the initial volume and temperature, and V₂ and T₂ are the final volume and temperature. PV = nRT is the formula for the ideal gas law. Where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, T is the temperature of the gas in Kelvin, and R is the universal gas constant.
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11. A man lifts a weight of 300 N through a
vertical height of 2 m in 6 seconds. What
power does he develop?
Explanation:
Weight=force=300N
height=2m
time=6seconds
power=?
now,
power (p)=f×h÷t
power=300×2÷6
power=100w
therefore,A man develops 100 watt power in 6 seconds.
weightless spring has a spring constant of 1.85 n/m. a 500-g mass is attached to the spring. it is then displaced 10.0 cm and released. find the total energy of the mass
The mass is displaced 10.0 cm and released. We will use the formula for the total energy of the mass.Total energy of a mass-energy system is the sum of potential energy and kinetic energy. Mathematically, it is given as:Etotal = Ep + Ekwhere,
Etotal = Total energy of the systemEp = Potential energy of the systemEk = Kinetic energy of the systemGiven,Mass of the system = 500 g = 0.5 kgSpring constant of the spring, k = 1.85 N/mDisplacement, x = 10.0 cm = 0.1 mWe can find the potential energy of the mass using the formula for potential energy stored in a spring as,Ep = (1/2)kx²Putting the values,Ep = (1/2) × 1.85 N/m × (0.1 m)²Ep = 0.00925 JThe mass is displaced and released, so the potential energy of the mass converts into kinetic energy at the maximum displacement.
Therefore, we can find the kinetic energy of the mass using the formula,Ek = (1/2)mv²where, v is the velocity of the mass at maximum displacement. We can find the velocity using the conservation of energy principle that is total energy of the system is equal to kinetic energy at the maximum displacement.Etotal = EkUsing the values,0.00925 J + Ek = EkEk = 0.00925 JThe kinetic energy is also given asEk = (1/2)mv²Putting the values,0.00925 J = (1/2) × 0.5 kg × v²v² = 0.0185 m²/s²v = 0.136 m/sNow, we can calculate the kinetic energyEk = (1/2)mv²Putting the values,Ek = (1/2) × 0.5 kg × (0.136 m/s)²Ek = 0.00587 JTherefore, the total energy of the mass attached to the spring is,Etotal = Ep + Ek= 0.00925 J + 0.00587 J= 0.01512 JHence, the total energy of the mass is 0.01512 J.
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A car drives 2 miles north, 4 miles west, 2 miles south and then drives 6 miles east. What is the car's distance?
What is the car's displacement in this same example? (2 answers) (Must show work)
Answer:
Ok, so the car has traversed a total of 2 + 4 + 2 + 6 = 14 miles.
That's the car's distance! :)
For the displacement, you can draw out a diagram. I'll try my best to make one using a keyboard xDDD
- - - - |
| |
| - - - start - end!
So, this is the journey of the car: Going two up, 4 left, 2 down, and 6 right.
This ends up 2 to the right of the beginning!
That's a 2 mile displacement :)
Can you give me brainliest for my splendid diagram? xD
