Given gravitational potential energy when he's lifted is 2058 J.
Kinetic energy is transferred to the person.
Amount of kinetic energy the person has is -2058 J
velocity of person = 7.67 m/s².
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Explanation:</h3>
Given:
Weight of person = 70 kg
Lifted height = 3 m
1. Gravitational potential energy of a lifted person is equal to the work done.

Gravitational potential energy is equal to 2058 Joules.
2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.
3. Kinetic energy gained = Potential energy lost = 
Kinetic energy gained by the person = (-2058 kg.m/s²)
4. Velocity = ?
Kinetic energy magnitude= 
Solving for v, we get

The person will be going at a speed of 7.67 m/s².
Answer:
The first graph is showing the constant acceleration (1 m/s)
Explanation:
The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4
The last graph is showing constant velocity therefore there is no acceleration (a = 0)
Answer:
The answer is D. right atrium, left atrium, right ventricle, left ventricle
Explanation:
The right atrium receives oxygen-poor blood from the body and pumps it to the right ventricle.The right ventricle pumps the oxygen-poor blood to the lungs.The left atrium receives oxygen-rich blood from the lungs and pumps it to the left ventricle.The left ventricle pumps the oxygen-rich blood to the body.
Answer:
The energy dissipated as the puck slides over the rough patch is 1.355 J
Explanation:
Given;
mass of the hockey puck, m = 0.159 kg
initial speed of the puck, u = 4.75 m/s
final speed of the puck, v = 2.35 m/s
The energy dissipated as the puck slides over the rough patch is given by;
ΔE = ¹/₂m(v² - u²)
ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)
ΔE = -1.355 J
the lost energy is 1.355 J
Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J
<h2>Given :</h2>
- total charge = 9.0 mC = 0.009 C
Each electron has a charge of :

For producing 1 Cuolomb charge we need :
Now, for producing 0.009 C of charge, the number of electrons required is :
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So, Number of electrons passing through the cross section in 3.6 seconds is :

Number of electrons passing through it in 1 Second is :
Now, in 10 seconds the number of electrons passing through it is :
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