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kompoz [17]
3 years ago
15

a 10 kilogram steel ball is dropped from the top of a tower 100 meters high the kinetic energy of the ball just before it sttike

s the ground is most nearly​
Physics
1 answer:
EleoNora [17]3 years ago
4 0

Explanation:

If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.

Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.

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A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the
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Answer:

Explanation:

  • The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
  • where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
  • The potential energy of the system is given as ; PE = mgh
  • The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
  • The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
  • Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

I = mR²

The moment of inertia of the ring is maximum and therefore reaches the bottom last.

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3 years ago
A transformer has 100 turns in its primary coil and 75 turns in its secondary coil. If the input voltage is 12.0 V, what is the
Sindrei [870]
Is 54 por que es igual
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2 years ago
A CD has to rotate under the readout-laser with a constant linear velocity of 1.25 m/s. If the laser is at a position 3.7 cm fro
Savatey [412]

Answer:N=322.53 rpm

Explanation:

Given

Linear velocity (v)=1.25 m/s

Position from center is 3.7 cm

we know

v=\omega \times r

1.25\times 100=\omega \times 3.7

\omega =\frac{125}{3.7}=33.78

and \frac{2\pi N}{60}=\omega

N=\frac{\omega \times 60}{2\pi }

N=\frac{33.78\times 60}{2\pi }

N=322.53 rpm

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What does it mean to be skeptical of health fraud
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An airplane is heading due south at a speed of 430 km/h . A wind begins blowing from the northwest at a speed of 85.0 km/h (aver
Igoryamba

Answer:

It should fly 8° to west of south at 430km/h

Explanation:

According to the diagram. X components for both velocities must have the same magnitude in order to get the resultant velocity due south.

V_{w}*cos(45) = V_{A}*sin(\alpha )   Solving for α:

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