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Fiesta28 [93]
3 years ago
9

g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positi

ve constant. Part A What is the force that one atom exerts on the other? Express your answer in terms of C6 and x. Fx = nothing Request Answer Part B Is this force attractive or repulsive? Is this force attractive or repulsive? attractive repulsive
Physics
1 answer:
Arisa [49]3 years ago
5 0

Answer:

F_x = -\frac{6 C_6}{2^7}

Attractive

Explanation:

Data provided in the question

The potential energy of a pair of hydrogen atoms given by \frac{C_6}{X_6}

Based on the given information, the force that one atom exerts on the other is

Potential energy μ = \frac{C_6}{X_6}

Force exerted by one atom upon another

F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X}  (-\frac{C_6}{X^6})

or

F_x = \frac{\partial}{\partial X}  (\frac{C_6}{X^6})

or

F_x = -\frac{6 C_6}{2^7}

As we can see that the C_6 comes in positive and constant which represents that the force is negative that means the force is attractive in nature

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How much time will elapse between seeing and hearing an event?
julia-pushkina [17]

Depends on how far away the event is and what the temperature is as this affects the speed of sound.

For example, let's say you're 600 meters away and the temperature has no affect.

The speed of sound would be roughly 340 m/s so the time it would take to hear the sound would be 600/340 = 1.76 seconds

The speed of light (c) is 3.0 X 10^8 m/s so the time it would take to see the event would be 600/3 X 10^8 = 2 X 10^-7

Subtract: 1.76 - (2 X 10^-7) = approx. 1.76

3 0
3 years ago
How can a model be created to predict how forces affect objects at a distance?
olchik [2.2K]

Answer:

Gravitational force increases as the masses of the objects increase and decreases as the distance between the objects increases. Balanced forces acting on an object cause no change in the motion of the object. When unbalanced forces act on an object, the sum of the forces is not equal to zero.

Explanation:

put it in your own words

8 0
3 years ago
A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha
Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss} (1)

From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that m = 175\,kg, g = 9.807\,\frac{m}{s^{2}}, h_{1} = 18\,m, h_{2} = 8\,m and v_{2} = 11\,\frac{m}{s}, then the work done by non-conservative force is:

W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0
2 years ago
A ball of 10kg falls from rest from a height of 150m, Neglating air resistance, calculate its kinetic energy after falling a dis
GaryK [48]

Answer: 3920\ J

Explanation:

Given

mass of ball m=10 kg

It is placed at a height of 150 m

It is dropped from the height and allowed to free fall for 40 m

Velocity acquired by the ball during this fall is given by v^2-u^2=2as

Insert u=0, a=g

\Rightarrow v^2-0=2\times 9.8\times 40\\\Rightarrow v=\sqrt{784}\\\Rightarrow v=28\ m/s

Kinetic energy at this instant

K.E.=\dfrac{1}{2}\times 10\times 28^2\\\\\Rightarrow K.E.=3920\ J

3 0
2 years ago
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