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GrogVix [38]
3 years ago
5

A 2-ton car is parked on a driveway. Which of these is the reaction force if the weight of the car is the action force? A. the m

ass of the car B. gravity pulling on the car C. the weight of the driveway D. the driveway pushing up on the car
Physics
2 answers:
tester [92]3 years ago
6 0

Answer:

The driveway pushing up on the car

Explanation:

A reaction force is the force that acts upward on an object resting on a particular surface to balance the effect of the weight of the object which acts downward. For this 2-ton car parked on a driveway, the reaction from the floor of the driveway acts upward on the car to balance the downward force of the weight of the car since in every action, there is equal and opposite reaction. By this force balancing, the car will neither move up nor sink down to the ground as long as it is parked on the driveway.

UNO [17]3 years ago
5 0
A. because mass can also deal with weight and the other choices don't make sense
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In this context, one feature that makes model A better is that this represents the molecule using a 3D model, which is better to understand how the molecule looks like and what is its structure. Moreover, both models are alike because they show the number of atoms of each element, although model A does not show the types of elements.

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The spring of modulus k = 200 n /m is compressed a distance of 300 mm and suddenly released with the system at rest. determine t
DiKsa [7]
I attached the missing picture.
Let's analyze the situation as spring goes from stretched to unstretched state.
When you stretch the string you have to use force against ( you are doing work) this energy is then stored in the spring in the form of potential energy. When we release the spring the energy is being used to push the two carts. When the spring reaches its unstretched length its whole initial potential energy has been used on the carts, and this is the moment when two carts have maximum velocity.
The potential energy of compressed ( stretched) spring is:
E_p=\frac{1}{2}kx^2
The kinetic energy of two carts is:
E_{k1}+E_{k2}=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}
So we have:
E_p=E_{k1}+E_{k2}\\ \frac{1}{2}kx^2=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}
Momentum also has to be conserved:
m_1v_1-m_2v_2=0\\ m_1v_1=m_2v_2\\ v_1=\frac{m_2}{m_1}v_2
Momentum before the release of the spring is zero so it has to stay zero. We plug this back into the expresion we got from law of conservation of energy and we get:
v_2^2=\frac{m_1^2}{m_2^2-m_1^2}kx^2=4.05\\
v_2=\sqrt{4.05}=2.012\frac{m}{s}
Now we go back to the momentum equation:
v_1=\frac{m_2}{m_1}v_2\\
v_1=4.69\frac{m}{s}

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