Answer:
Their efforts would be expressed in units of Joules per second
Explanation:
The unit of their efforts can be derived from the formula of power which is given by the product of mass, acceleration and distance (the product is energy with unit joules) divided by time taken to complete the task (unit is seconds)
Therefore, the unit of their efforts would be joules per second
Answer:
d
Explanation:
there is not enough information about the liquid to know the force required for each. ie. stirring a cup of water is different than stirring a cup of pudding.
Answer:
You wouldnt fall you would be sucked and you would lose all air supply and you lungs would pop
Explanation:
Answer:

Explanation:
First of all, we need to find the pressure exerted on the sphere, which is given by:

where
is the atmospheric pressure
is the water density
is the gravitational acceleration
is the depth
Substituting,

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m
So the total area of the sphere is

And so, the inward force exerted on it is

Answer: 1026s, 17.1m
Explanation:
Given
COP of heat pump = 3.15
Mass of air, m = 1500kg
Initial temperature, T1 = 7°C
Final temperature, T2 = 22°C
Power of the heat pump, W = 5kW
The amount of heat needed to increase temperature in the house,
Q = mcΔT
Q = 1500 * 0.718 * (22 - 7)
Q = 1077 * 15
Q = 16155
Rate at which heat is supplied to the house is
Q' = COP * W
Q' = 3.15 * 5
Q' = 15.75
Time required to raise the temperature is
Δt = Q/Q'
Δt = 16155 / 15.75
Δt = 1025.7 s
Δt ~ 1026 s
Δt ~ 17.1 min