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Alona [7]
3 years ago
8

The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,

during this trip? Would the angular displacement be more or less if you put on big mud tires with a radius of .60m Mud tires have (more or less) angular displacemen
Physics
1 answer:
Travka [436]3 years ago
8 0
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
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Answer:

Torque decreases .

Explanation:

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Reel is in the form of disc

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3 years ago
During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive directio
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Answer:

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

\alpha = a / r = 7.6 / 0.26 = 29.23 rad/s^2

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where v = 0 m/s is the final angular velocity of the wheel when it stops, \omega_0 = 95rad/s is the initial angular velocity of the wheel, \alpha = -29.23 rad/s^2 is the deceleration of the wheel, and \Delta \theta is the angle swept in rad, which we care looking for:

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\Delta \theta = 9025 / 58.46 = 154.375 rad

As each revolution equals to 2π, the total revolution it makes before stop is

154.375 / 2π = 24.57 revolutions

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a skinny girl devours food that is equivalent to 600 kcal, as she is anorexic she repents and wants to lose twice as many calori
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Answer:

Explanation:

Work done in lifting the weight once = mgh

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Total energy to be spent = 600 x 10³ cals

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= 3.57  x 10³ times

Total time to be taken = 2 x 3.57 x 10³

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