a) in the upper position. b) in the lower position. c) in the lower position. d) in the upper position. f) Its kinetic and potential energy will be 0, but the energy is transferred to the element or body that stopped the movement of the pendulum

Explanation:

In the attached image we have the sketch of a pendulum system.

A) The potential energy is maximum when the pendulum is in the upper position (image, fig 1) because the elevation (h) is maximum with respect to the reference point.

B) the potential energy is minimum when the pendulum is in the lower pasition (image, fig 2) because the elevation (h) is cero with respect to the reference point.

Note: When the pendulum is coming down the potential energy is transforming in kinetic energy.

C) The kinetic energy is maximum when the pendulum is in the lower position (image, fig 2), because the potential energy has been transformed in kinetic energy.

D) The kinetic energy is maximum when the pendulum is in the upper position (image, fig 1) because at this moment the pendulum is at rest it means its velocity is 0. We know that the kinetic energy depends on the velocity.

f) The energy is transferred to the element or body that stopped the movement of the pendulum

4 0

2 years ago

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

6 0

3 years ago