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3 years ago

6. Draw a velocity-time graph for an object originally traveling at -3 m/s. The object

Physics

1 answer:

faltersainse [42]3 years ago

4 0

See the graph in attachment

Explanation:

In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.

In the graph, we see that the initial velocity at time t = 0 is

$v_0 = -3 m/s$

and it is negative, so below the x-axis.

Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).

After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.

Learn more about velocity:

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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

4 years ago

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A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

nignag [31]

Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block. The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that

[tex]f=mgsin \theta [tex]

Therefore, option B is the correct option.

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valentina_108 [34]

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4 years ago

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Monica [59]

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8_murik_8 [283]

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