Please help me im struggling

a child is swinging on swing, describe what happens to both the kinectic energy and potential enegry of the child as she swings

Igoryamba

Answer:

The K.E is maximum when the child is at the vertical position and the P.E is maximum at the extreme deviated position from the vertical.

Explanation:

A child is swinging on swing up and down has both kinetic and potential energy.
The total mechanical energy of the system is conserved throughout the system. At any instant the total mechanical energy is given by,

E = K.E + P.E

The K.E is maximum when the child is at the vertical position.
The P.E is maximum at the extreme deviated position from the vertical.
And when K.E is maximum P.E becomes minimum and vice versa as per the law of conservation of energy.

5 0

3 years ago

A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the

11Alexandr11 [23.1K]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\$

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

$(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\$

4 0

2 years ago

A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan

Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

$E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C$

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

$E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C$

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0

3 years ago

All matter is made of tiny particles which are in constant...

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Which are in constant motion.

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3 years ago

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