Answer:
Yes, it could be opened
Explanation:
Pressure is force per unit area

Since there is a difference in air pressure on both sides, this difference when multiplied by the area of the floor will create a force which could open the door, if this force is complimented with a greater applied force.
Answer:
9.1m/s
Explanation:
Nate throws a straight ball to Kayla who is standing at a balcony 3.8m above Nate
When she catches the ball, it is still moving upward with a speed of 2.8m/s
v = 2.8m/s
u = ?
s = 3.8m
a= -9.8(The acceleration has a negative sign because the speed of the ball is declining)
Therefore the initial speed at which Nate threw the ball can be calculated as follows
v^2= u^2 + 2as
2.8^2= u^2 + 2(-9.8)(3.8)
7.84= u^2 + (-74.48)
7.84= u^2 - 74.48
u^2= 7.84 + 74.48
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u^2= 82.32
u= √82.32
u = 9.1m/s
Hence the initial speed at which Nate threw the ball is 9.1m/s
U need to write better u egg head kid or tell your mom
Answer:
A) F = - 8.5 10² N, B) I = 21 N s
Explanation:
A) We can solve this problem using the relationship of momentum and momentum
I = Δp
in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction
v₀ = 8.50 m / s
v_f = -8.50 m / s
F t = m v_f -m v₀
F =
let's calculate
F =
F = - 8.5 10² N
B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations
v² = v₀² - 2g (y- y₀)
as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0
v =
calculate
v =
v = 14 m / s
to calculate the momentum we use
I = Δp
I = m v_f - mv₀
when it hits the ground its speed drops to zero
we substitute
I = 1.50 (0-14)
I = -21 N s
the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is
I = 21 N s