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laila [671]
3 years ago
10

Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they

Physics
1 answer:
Svetradugi [14.3K]3 years ago
6 0

Answer: Graph C is the correct option

Explanation:

The question is incomplete, please remember to submit the whole question :)

However, the rest of the question with its corresponding figures is below:

Assume that positive momentum is to the right, which of the following best represents the momentum of the cart of mass m as a function of time before and after the collision?

The initial momentum p_{o} of the cart with mass m (before the collision) is:

p_{o}=-mV (1) Note the negative sign indicates the direction of cart's velocity (to the left, as seen in the first image attached)

On the other hand, the final momentum p_{f} of both carts (after the inelastic collision) is:

p_{f}=(2m+m) V_{f}=3m V_{f} (2)

So, according to this, the correct graph tha best represents the situation is C. Since before the collision the momentum is negative, then both carts slow down after the collision (V_{f}), and taking into account the linear momentum is directly proportional to the velocity p_{f} (although is in the positive direction) is less than p_{o}.

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A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Ω resistor, and an open switch.A 8.2-V battery is connected
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Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance  in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})

Put the value into the formula

I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})

I(t)=0.01925\ A

I(t) = 19.25\ mA

(B). We need to calculate the store energy in the inductor

Using formula of energy

E=\dfrac{1}{2}LI^2

Put the value into the formula

E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2

E=7.04\times10^{-6}\ J

{tex]E=7.04\ \mu J[/tex]

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