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leva [86]
1 year ago
15

21. A 1000-kg automobile moving with a speed of 24 m/s relative to the road collides with a 500-kg automobile initially at rest.

If the two stick together, what is the velocity in m/s of the two cars after the collision according to an observer in a truck moving 10 m/s in the same direction as the moving cars
Physics
1 answer:
Luda [366]1 year ago
4 0

A truck is moving with less velocity in the direction in which the truck is moving earlier because the truck has more momentum.

<h3 /><h3>In which direction the truck moves?</h3>

A truck is moving with the velocity of 10 m/s in the same direction in which the truck is moving earlier because the truck has more mass so it has more momentum. Due to collision, the velocity of the truck is slow down but can't be stopped because of high momentum in the truck.

So we can conclude that a truck is moving with less velocity in the direction in which the truck is moving earlier because the truck has more momentum.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

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Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.

Explanation:

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