Answer:
it would take 3.26 seg for the stone to fall to the water
Explanation:
If we ignore air friction then:
h=h₀ + v₀*t -1/2*g*t²
where
h= coordinates of the stone in the y axis ( height of the stone relative to the surface of the water )
h₀ = initial coordinates of the stone ( height of the cliff relative to the surface of the water = 52 m )
v₀ = initial <u>vertical </u>velocity = 0 ( since the ball is kicked horizontally , has only initial horizontal velocity , and has 0 vertical velocity )
t = time to reach a height h
g = gravity = 9.8 m/s²
since v₀ =0
h= h₀ - 1/2*g*t²
h₀ - h = 1/2*g*t²
t= √[2(h₀ - h)/g]
when the stone hits the ground h=0 ( height=0) , then replacing values
t=√[2(h₀ - h)/g]=√[2(52 m- 0 m )/(9.8m/s²)] = 3.26 seg
t= 3.26 seg
it would take 3.26 seg for the stone to fall to the water
Answer:
50J
Explanation:
At the top you have(A)
KE_a = O
PE_a = 100J
KE + PE = 100J
At the bottom you have (C)
KE_c= 100J
PE_c=0J
KE+PE = 100J
At point C:
You are at half the height.
We know that at H, PE =100J
PE_c = mgH
At C,
PE_c= mg (H/2) *at half the height
*m and g stay the same
Intuitively, the higher you are, the more potential energy you have.
If you decrease the height by a half, your PE will also decrease
At A:
PE_a / (mg) = H
At B:
PE_b / (mg) = H/2
to also get H on the right hand side, multiply by 2
2 (PE_b/ (mg))= H
2PE_b / (mg) = H
Ok, now that we have set up 2 equations (where H is isolated), find PE at B
AT A = AT B *This way you are saying that H = H (you compare both equations)
PE_a / (mg) = 2x PE_b / (mg)
*mg are the same for both cancel them (you can do that because of the = sign)
PE_a = 2PE_b
We know that PE_a = 100J
100J/2 = PE_b
PE at b = 50J
**FIND KE at b
We know that
KE_b + PE_b is always 100J
100J = 50J + KE_b
KE_b = 50J
More energy, colour and I think less bright
Answer: Option D: 5.5×10²Joules
Explanation:
Work done is the product of applied force and displacement of the object in the direction of force.
W = F.s = F s cosθ
It is given that the force applied is, F = 55 N
The displacement in the direction of force, s = 10 m
The angle between force and displacement, θ = 0°
Thus, work done on the object:
W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J
Hence, the correct option is D.
Answer:
Between the principal focus and the pole of the mirror
