Answer:
Explanation:
Let the height of the tree is y and the distance of tree from point B is x.
According to the diagram

x = 0.55 y ..... (1)

(50 + 0.55y) 1.17 = y ..... from equation (1)
58.5 + 0.644 y = y
0.356 y = 58.5
y = 164.3 ft
Answer:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. The total momentum of the trolleys after separation is zero
c. The momentum of the 2 kg trolley after separation is 12 kg·m/s
d. The momentum of the 3 kg trolley is -12 kg·m/s
e. The velocity of the 3 kg trolley = -4 m/s
Explanation:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0
c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s
d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s
e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley
∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s
The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s
Use formula, v^2= u^2 + 2as.
The "v" and the "s" of the formula are given.
Since u is 0, just use f=ma.
I hope this helped!
Answer:

Explanation:
Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.
Horizontal distance = 13 m
Vertical distance = 57 cm
Lets take time to cover 57 cm distance in vertical direction is t.
We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

Here 
S= 57 cm

t=0.34 s
Now in the horizontal direction

Here x=13 m
t= 0.34 s
So


So the initial speed of ball is 38.13 m/s.
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3