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Katena32 [7]
3 years ago
11

Some curious students hold a rolling race by rolling four items down a steep hill. The four items are a solid homogeneous sphere

, a thin spherical shell, a solid homogeneous cylinder and a hoop with all its mass concentrated on the hoop's perimeter. All of the objects have the same mass and start from rest. Assume that the objects roll without slipping and that air resistance and rolling resistance are negligible. For each statement below, select True or False.
a) Upon reaching the bottom of the hill, the hoop will have a larger rotational kinetic energy than any of the other objects will when they reach the bottom of the hill.
A: True B: False

b) The hoop reaches the bottom of the hill before the homogeneous cylinder.
A: True B: False

Physics
1 answer:
Papessa [141]3 years ago
8 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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Suppose that Lisa walks her dog around the block for a little exercise. The block is 1 mile around. If she walks around the bloc
natita [175]
Displacement only measure how far between the starting and ending point. In this case, Lisa walks around the block as a circle so the starting point is the same as the ending point. Thus, displacement is 0mile.
On the other hand, distance measures exactly how far she walks. In this case, the distance is 1 mile, same as the perimeter of the block.
7 0
3 years ago
Air (14.5 lb) undergoes a polytropic process in a closed system from p1 = 80 lbf/in2, υ1 = 4 ft3/lb to a final state where p2 =
Yanka [14]
The energy transfer in terms of work has the equation:

W = mΔ(PV)

To be consistent with units, let's convert them first as follows:

P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm

W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf

In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>


4 0
3 years ago
A man is flying in a hot-air balloon in a straight line at a constant rate of 5 feet per second, while keeping it at a constant
icang [17]

Answer:

x = 220.85 ft

Explanation:

Let at any moment of time the friend's car is at some horizontal distance "x" from the position of balloon.

Now if the altitude of the balloon is fixed and it is at height "h"

so here we will have

tan \theta = \frac{h}{x}

now we know that

initially the angle of the friend's car is 35 degree

so the horizontal distance will be

x_1 = h cot35

similarly if the angle after passing the car position is 36 degree

then we have

x_2 = h cot36

now the speed of the balloon is constant

so we have

v = \frac{x_1 + x_2}{\Delta t}

5 ft/s = \frac{h cot35 + h cot36}{90 s}

5 ft/s = \frac{2.8h}{90}

h = 160.45 ft

so the final position of friend when the angle is 36 degree

x = \frac{h}{tan36}

x = \frac{160.45}{tan36}

x = 220.85 ft

4 0
3 years ago
A rocket accelerates at 25m/s^2 for 5s before it runs out of fuel and dies. What is the maximum height reached by the rocket?
sineoko [7]

Answer:

1109m

Explanation:

The distance h₁ traveled by the rocket during acceleration:

h_1=\frac{1}{2}at^2

The velocity v₁ at this height h₁:

v_1=at

When the rocket runs out of fuel, energy is conserved:

E=\frac{1}{2}mv_1^2+mgh_1=mgh_2

Solving for h₂:

h_2=\frac{v_1^2}{2g}+h_1=\frac{(at)^2}{2g}+\frac{1}{2}at^2

8 0
3 years ago
BRAINLIEST PLEASE: You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes yo
Sophie [7]

Answer:

The velocity with which the stone strikes the water surface is 33.0 m/s (nearest tenth)

Explanation:

Good luck!

3 0
2 years ago
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