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3 years ago

An 80.0 kg skier slides down a hill shaped as shown. Assume

Physics

1 answer:

umka21 [38]3 years ago

8 0

The height above the ground from where the skier start is 11.5 m.

<h3>Conservation of energy</h3>

The height above the ground from where the skier start is determined by applying the principle of conservation of energy as shown below;

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

$h = \frac{v^2}{2g} \\\\h = \frac{15^2}{2 \times 9.8} \\\\h = 11.5 \ m$

Thus, the height above the ground from where the skier start is 11.5 m.

Learn more about conservation of energy here: brainly.com/question/166559

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The velocity is 1100 kilometers per hour

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3 years ago

If you were to be drawn into a black hole, what would happen? To the black hole, not to you.

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3 years ago

A thin ball shell is shaped like a hollow ball with a radius of 42.8 cm. The material is a conductor with charge 0.8 nC A point

Gala2k [10]

Answer:

The Electric field will be 225.92 N/C

Explanation:

Given :

Radius of the hollow sphere R=42.8 cm
Initial charge on the conducting sphere is $Q_1=0.8\times10^{-9}\ \rm C$
Magnitude of the point charge $q_2=-5.4\times10^{-9}\ \rm C$

We know that the flux of the electric field through a close surface cab be calculated by using Gauss Law which is

$\int EdA=\dfrac{Q_{in}}{\epsilon_0}\\\\E\times4\pi\times 0.428^2=\dfrac{(0.8-5.4)\times10^{-9}}{\epsilon_0}\\\\E=225.92\ \rm N/C$

Hence the Electric Field is calculated.

The answer founded out by you seems to be correct and very close to the exact answer and the concept you have used in your answer is also correct.

5 0

3 years ago

An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindston

Dmitry_Shevchenko [17]

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

6 0

4 years ago