Answer:
Explanation:
specific gravity of oil is 
we know that
change in pressure for oil is given as
here density and h is for oil
change in pressure for WATER is given as
here density is for water and h is for water
pressure change due to both is given as
Answer:
2.26 s
Explanation:
Let's take down to be positive.
Given (in the y direction):
Δy = 25 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
25 m = (0 m/s) t + ½ (9.8 m/s²) t²
25 = 4.9t²
t = 2.26 s
If the ball instead had an initial horizontal velocity of 5 m/s, its initial vertical velocity is still 0 m/s. So the time to fall is still 2.26 s.
I believe the answer is a
Answer:
v₀ₓ = 63.5 m/s
v₀y = 54.2 m/s
Explanation:
First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:
K.E = (0.5)(mv₀²)
where,
K.E = initial kinetic energy of projectile = 1430 J
m = mass of projectile = 0.41 kg
v₀ = launch velocity of projectile = ?
Therefore,
1430 J = (0.5)(0.41)v₀²
v₀ = √(6975.6 m²/s²)
v₀ = 83.5 m/s
Now, we find the launching angle, by using formula for maximum height of projectile:
h = v₀² Sin²θ/2g
where,
h = height of projectile = 150 m
g = 9.8 m/s²
θ = launch angle
Therefore,
150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)
Sin θ = √(0.4216)
θ = Sin⁻¹ (0.6493)
θ = 40.5°
Now, we find the components of launch velocity:
x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)
<u>v₀ₓ = 63.5 m/s</u>
y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)
<u>v₀y = 54.2 m/s</u>
Answer:
Explanation:
Given
mass 
Force 
door knob is located at a distance of r=0.8 m from axis
Angular acceleration of door 
Torque 
where I=moment of inertia
