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Brums [2.3K]
3 years ago
15

A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho

uld a solid sphere of the same mass and radius be released to have the same speed as the cylinder at the bottom of the hill?
Physics
1 answer:
Jlenok [28]3 years ago
3 0

Answer:

same 0.81m

Explanation:

in this problem if we assume there no resistance of any sort. and we apply the energy conservation

change in Potential energy = change in kinetic energy

mgh = 0.5mv^2

gh = 0.5v^2

the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

so at the bottom

put h = 0.81m

9.81 * 0.81 * 2 = v^2

v=3.99 m/s

both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size

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A 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. At one moment it is moving 5.0 m/s. A little l
Y_Kistochka [10]
The change in kinetic energy of the car is equivalent to the change in its potential energy. Thus:
K.E  = P.E
1/2 x mΔv² = mgΔh
h = (8.2² - 5²) / 2(9.81)
h = 2.15 meters
4 0
4 years ago
Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height require
strojnjashka [21]

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

       Em₀ = U = m g h

Final point. In the highest part of the loop

       Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

      I = m R²

      v = w R

we substitute

       Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

       em_f = m v² + 2 m g R

Energy is conserved

       Emo = Em_f

       mgh = m v² + 2m g R

       h = v² / g + 2R

 

The lowest velocity that the ball can have at the top of the loop is v> 0

      h> 2R

3 0
3 years ago
A box is at rest on a table. What can you say about the forces acting on the box?
GarryVolchara [31]

the upward force is from the table and the downward is from gravity they are equal in force so the box doesn't fly up or sink down


8 0
3 years ago
Read 2 more answers
The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei
Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+6450} \right )^2

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+33700} \right )^2

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0
3 years ago
In an elastic collision, _______ energy is conserved.
Nonamiya [84]

Answer:

Kinetic energy and momentum are conserved.

Explanation:

4 0
2 years ago
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