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alukav5142 [94]
3 years ago
11

A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

4.58 Correct: Your answer is correct. rad/s (b) Determine the tangential speed at a point 2.98 cm from its center. 3.71 Correct: Your answer is correct. m/s (c) Determine the radial acceleration of a point on the rim. magnitude 1.23 Correct: Your answer is correct. km/s2 direction toward the center Correct: Your answer is correct. (d) Determine the total distance a point on the rim moves in 2.06 s. 20.28 Correct: Your answer is correct. m
Physics
1 answer:
Tanya [424]3 years ago
8 0

Answer:

124.62\ \text{rad/s}

3.71\ \text{m/s}

1.23\ \text{km/s}^2

20.28\ \text{m}

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}

The angular speed is 124.62\ \text{rad/s}

r = 2.98 cm

Tangential speed is given by

v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}

Tangential speed at the required point is 3.71\ \text{m/s}

Radial acceleration is given by

a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2

The radial acceleration is 1.23\ \text{km/s}^2.

t = Time = 2.06 s

Distance traveled is given by

d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}

The total distance a point on the rim moves in the required time is 20.28\ \text{m}.

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Answer:

Following are the answer to this question:

Explanation:

In option (a):

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  • Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.

In option (b):

  • Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.  
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3 years ago
Lithium (chemical symbol Li) is located in Group 1, Period 2. Which is lithium most likely to be? O A. A soft, shiny, highly rea
algol13
A. A soft, shiny, highly reactive metal
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A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr
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Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds

The reaction time cannot be more than 1.69515 seconds

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Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur
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Answer:

-1.7908787542\times 10^{-9}\ C

3.4260289211\times 10^{-9}\ C

1.7908787542\times 10^{-9}\ C

1.6351501669\times 10^{-9}\ C

Explanation:

r = Radius

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electric field is given by

E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C

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Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C

The charge is 3.4260289211\times 10^{-9}\ C

The charge inside will have the polarity changed

q=+1.7908787542\times 10^{-9}\ C

Outside the charge will be

3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C

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Imani stirs a cup of hot sencha tea with a cool silver spoon. She notices that the silver spoon becomes warmer. What energy chan
Lina20 [59]

Answer:

D. The tea loses heat to the spoon causing the spoon to become warmer

Explanation:

When the silver spoon at a lower temperature than the tea, is added to the tea, it makes thermal contact. Hence, the heat transfer starts between the two until the equilibrium is reached. We know that the heat transfer takes place from the body with a higher temperature to a body with a lower temperature. As a result, the body with higher temperature loses heat and its temperature lowers down. While the body with a lower temperature gains heat and its temperature rises.

Therefore, the correct option is:

<u>D. The tea loses heat to the spoon causing the spoon to become warmer</u>

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