Answer:
A) To true. he pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases
Explanation:
Let us propose the solution of this problem before seeing the final statements. The pressure increases with the depth of raposin due to the weight of water that is above the person and also the pressure exerted by the atmosphere on the entire pool, the equation describing this process is
P =
+ ρ g y
Where is the atmospheric pressure, ρ the water density, and 'y' the depth measured from the surface.
Let's examine this equation in we see that the total pressure is directly proportional to the atmospheric pressure and depth
Now we can examine the claims
A) To true. State agreement or with the equation above
B) False. Pressure changes with atmospheric pressure
C) False. It's the opposite
D) False. They are directly proportional
Answer:
a) W_total = 8240 J
, b) W₁ / W₂ = 1.1
Explanation:
In this exercise you are asked to calculate the work that is defined by
W = F. dy
As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.
W = F dy = F Δy
let's apply this formula to our case
a) Let's use Newton's second law to calculate the force in the first y = 5 m
F - W = m a
W = mg
F = m (a + g)
F = 80 (1 + 9.8)
F = 864 N
The work of this force we will call it W1
We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)
F₂ - W = 0
F₂ = W
F₂ = 80 9.8
F₂ = 784 N
The work of this fura we will call them W2
The total work is
W_total = W₁ + W₂
W_total = (F + F₂) y
W_total = (864 + 784) 5
W_total = 8240 J
b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use
W₁ / W₂ = F y / F₂ y
W₁ / W₂ = 864/784
W₁ / W₂ = 1.1
Answer:
Energy needed = 1100 kJ
Explanation:
Energy needed = Change in kinetic energy
Initial velocity = 15 m/s
Mass, m = 1600 kg
Final velocity = 40 m/s
Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J
Energy needed = 1100 kJ
Fm=Fe and am>ae
Hopefully this helps
