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Svetradugi [14.3K]
3 years ago
15

A wheel with a weight of 388 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it i

s rotating at an angular velocity of 23.6 rad/s . The radius of the wheel is 0.591 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of h above the bottom of the hill; this work has a magnitude of 3482 J .
Calculate h. Use 9.81m/s2 for the acceleration due to gravity.
Physics
1 answer:
Alisiya [41]3 years ago
4 0

the correct answer is 14m

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nataly862011 [7]
I think that it’s false I might be wrong but I want the points
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The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m
Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

\theta=\frac{s}{r}

For the Moon

\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad

The angle subtended by the Moon is 0.00903 rad

For the Sun

\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}

The area ratio is 6.268\times 10^{-6}

3 0
3 years ago
What are the thee major wind systems
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How much work does it take to move a 50 μC charge<br> against a 12 V potential difference?
lukranit [14]
<span>work =V*Q =12*50*10^-6

The total work done will be equal to 

work = V.Q

which means 

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>
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