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8_murik_8 [283]
3 years ago
9

A solid cylinder and a thin-walled hollow cylinder have the same radii. They are rolling horizontally toward the bottom of an in

cline. Assuming there is no friction, what is the maximum height each one reaches (in terms of the variables involved)?
Physics
1 answer:
Wittaler [7]3 years ago
8 0

Answer:

height solid / height hollow = 3/4

Explanation:

the height reached is directly proportional to the original KE of the object.

Total KE solid

=1/2mv^2+1/2\times1/2m r^2(v^2 / r^2 )=3/4mv^2

Total KE hollow = 1/2mv^2+1/2mr^2(v^2 / r^2 )=mv^2

height solid / height hollow = 3/4

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emmasim [6.3K]

Hi,

Newton's Law of Inertia explains this action because the resistance of the seat belt allows the change in its state of motion. So because the change of speed, your body moves forward but is restricted by the seat belt.

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A force of 50 N stretches a string by 4 cm,calculate the elastic constant.
murzikaleks [220]

Answer:

50/0.04= answer

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A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
4 years ago
A rifle of mass M is initially at rest, but is free to recoil. It fires a bullet of mass m with a velocity +v relative to the gr
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Answer:

V=-\dfrac{mv}{M}

Explanation:

Given that

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velocity of bullet =  v

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There is no any external force ,that is why linear momentum of the system will be conserve.

Initial linear momentum = Final  linear momentum

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0 =  M x V + m v

V=-\dfrac{mv}{M}

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Pavel [41]
I think it’s parallel
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