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3 years ago

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A mouse ran 25.45 meters in 11 seconds, stopped for 5 seconds to eat a piece of cheese, ran 2.5 meters in 1.2 seconds to hide be

hind a shelf, and finally ran another 9 seconds a distance of 15 meters to its home. Calculate the average speed of the mouse in m/s.

1 answer:

Nookie1986 [14]3 years ago

3 0

Okie so I can see it on tomorrow morning lol lol

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If a car travels 60 mph for a distance of 180 miles, how much time<br> did it take?

jolli1 [7]

Answer:

3 hours

Explanation:

180 divided by 60 (mph means miles per hours by the way)

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3 years ago

A car with a mass of 2.0 * 10^3 kg is traveling at 15 m/s. what is the momentum of the car?

Allushta [10]

Hello,

Your answer to this problem is 400/3

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2 years ago

Alternative sources of energy include : geothermal

baherus [9]

Answer:

All of the above

Explanation:

The correct answer is option E (All of the above)

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3 years ago

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

3 years ago

Read 2 more answers

Find the electric potential VP at point P. [Hint: To input a natural logarithm into the answer box, simply type the letters "ln"

Alik [6]

Answer:

After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+ $\frac{L}{d}$ )

Explanation:

I believe it is a part C question.

The derivative of V and P will be directly proportional to the differential dq and the inverse of Чπϵ₀δ........

Please find detailed solution in the attached picture as i believe that is the answer to the part C question you are seeking for.

8 0

3 years ago