Answer:
Explanation:
It is given that,
Depth of Death valley is 85 m below sea level, 
The summit of nearby Mt. Whitney has an elevation of 4420 m, 
Mass of the hiker, m = 65 kg
We need to find the change in potential energy. It is given by :
or
So, the change in potential energy of the hiker is 
. Hence, this is the required solution.
Answer:
I should be active for 15 hours to meet the physical activity requirement.
Explanation:
Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.
So, t = t₀/√(1 - β²)
t = 9/√(1 - (v/c)²)
= 9/√(1 - (0.8c/c)²)
= 9/√(1 - (0.8)²)
= 9/√(1 - (0.64)
= 9/√0.36
= 9/0.6
= 15 hr
So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.
As per the question the mass of the boy is 50 kg.
The boy sits on a chair.
We are asked to calculate the force exerted by the boy on the chair at sea level.
The force exerted by boy on the chair while sitting on it is nothing else except the force of gravity of earth i.e the weight of the body .The direction of that force is vertically downward.
At sea level the acceleration due to gravity g = 9.8 m/s^2
Hence the weight of the boy 
[m is the mass of the body]
we have m = 50 kg.
Hence w = 50 kg ×9.8 m/s^2
=490 N kg m/s^2
= 490 N
Here newton [N] is the unit of force.
Answer:
The natural medium emanating from the Sun and other very hot sources (now recognised as electromagnetic radiation with a wavelength of 400-750 nm), within which vision is possible.
Explanation:
just the way it is
Answer. Second Option: .85p_o=p_o e^-.00012h
Solution:
P(h)=Po e^(-0.00012h)
Air pressure: P(h)
Height above the surface of the Earth (in meters): h
Air pressure at the sea level: Po
Height at which air pressure is 85% of the air pressure at sea level:
h=?, P(h)=85% Po
P(h)=(85/100) Po
P(h)=0.85 Po
Replacing P(h) by 0.85 Po in the formula above:
P(h)=Po e^(-0.00012h)
0.85 Po = Po e^(-0.00012h)
