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masya89 [10]
3 years ago
5

. A 2.00 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.00 m, and th

e height of the room is 3.00 m. What is the gravitational potential energy associated with the ball relative to each of the following? (a) the ceiling (b) the floor (c) a point at the same elevation as the ball
Physics
1 answer:
cupoosta [38]3 years ago
4 0
<h2>a) Potential energy of ball relative to ceiling is 19.62 J</h2><h2>b) Potential energy of ball relative to floor is 39.24 J</h2><h2>c) Potential energy of ball relative to same elevation is 0 J</h2>

Explanation:

Mass of ball, m = 2 kg

Acceleration due to gravity, g = 9.81 m/s²

Height of ball from ground = 3 - 1 = 2 m

Potential energy, PE = mgh

Potential energy of ball, PE = 2 x 9.81 x 2 = 39.24 J

a) Potential energy of ball at ceiling = 2 x 9.81 x 3 = 58.86 J

Potential energy of ball relative to ceiling = 58.86 - 39.24 = 19.62 J

b) Potential energy of ball at floor = 2 x 9.81 x 0 = 0 J

Potential energy of ball relative to floor = 39.24 - 0 = 39.24 J

c) Potential energy of ball at same elevation = 2 x 9.81 x 2 = 39.24 J

Potential energy of ball relative to same elevation = 39.24 - 39.24 = 0 J

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Answer:

Explanation:

Given

mass of children m_1=20\ kg

m_2=30\ kg

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7 0
3 years ago
to move a resting box of 100 Newton on the ground with kinetic friction coeficient of 0,250 is applied a force of 60 N horizonta
krok68 [10]
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
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the work done is,
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The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J. 
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3 years ago
Depending on circumstances, there are times when divers may wish to be positively, negatively or neutrally buoyant. True. False.
ser-zykov [4K]

Answer:

True

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3 years ago
The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:
Furkat [3]
To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is  s=r(B), that has its <span>r=radius,B=angle in rad velocity v=ds/dt= w(r)
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Now we can go to proportion
v1=v2
w1*r1 = w2r2w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)
w2 = w3 (which is the   angular velocity of the rear wheel) &#10;
SOLVING FOR A : v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec
v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)= 9.99 or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!
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