Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine 
= 62 MPa = 62 x 10^6 Pa
we also know that = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine 
= 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure , we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>
Answer:
i think the answer is constant
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.
<h3>
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:
A) d_o = 20.7 cm
B) h_i = 1.014 m
Explanation:
A) To solve this, we will use the lens equation formula;
1/f = 1/d_o + 1/d_i
Where;
f is focal Length = 20 cm = 0.2
d_o is object distance
d_i is image distance = 6m
1/0.2 = 1/d_o + 1/6
1/d_o = 1/0.2 - 1/6
1/d_o = 4.8333
d_o = 1/4.8333
d_o = 0.207 m
d_o = 20.7 cm
B) to solve this, we will use the magnification equation;
M = h_i/h_o = d_i/d_o
Where;
h_o = 3.5 cm = 0.035 m
d_i = 6 m
d_o = 20.7 cm = 0.207 m
Thus;
h_i = (6/0.207) × 0.035
h_i = 1.014 m
Answer:
True
Explanation:
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