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Alex777 [14]
3 years ago
10

A car accelerates from a resting position at a constant 5 meters per second per second. What does the graph of acceleration vers

us time look like

Physics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0

You just told us that the acceleration is constant, and we believed you. So if we draw a graph of it, and it never gets any higher or lower than 5 m/s², then the graph is a straight horizontal line.

kvv77 [185]3 years ago
4 0

Explanation:

It is given that, a car accelerates from a resting position at a constant 5 meters per second per second. When the car moves with constant velocity, then the change in velocity is equal to 0.

The attached figure shows the acceleration time graph when the car accelerates with constant velocity. The line is parallel to the time axis or x axis.

So, the acceleration versus time graph look like line parallel to x axis. Hence, this is the required solution.

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Answer:

make it go faster

Explanation:

because of the arrow danmaicts of the force the wind give more speed

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In a solution, the solvent is present in the larger amount. <br> A True <br> B False
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To practice Tactics Box 9.1 Calculating the Work Done by a Constant Force. Recall that the work W done by a constant force F⃗ at
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Answer:

The vector magnitudes F and r are always postive, so the sign o W is determined entirely by the angle e between the force and the displacement.Submit Figure 1 off 1 part C

3 0
3 years ago
Read 2 more answers
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).

7 0
3 years ago
A ball of mass 0.500 kg is carefully balanced on a shelf that is 2.70 m above the ground. What is its gravitational potential en
ratelena [41]

Answer:

The gravitational potential energy of the ball is 13.23 J.

Explanation:

Given;

mass of the ball, m = 0.5 kg

height of the shelf, h = 2.7 m

The gravitational potential energy is given by;

P.E = mgh

where;

m is mass of the ball

g is acceleration due to gravity = 9.8 m/s²

h is height of the ball

Substitute the givens and solve for gravitational potential energy;

PE = (0.5 x 9.8 x 2.7)

P.E = 13.23 J

Therefore, the gravitational potential energy of the ball is 13.23 J.

8 0
3 years ago
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