Great question the answer is -25x.
Answer:
the rate of the change of the length of the shadow is - 0.8625 m/s.
The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.
Explanation:
Given the data in the question;
Let x represent the man's distance from building,
initially x = 1m2
dx/d t= -2.3 m/s
Also Let y represent shadow height
so we determine dy/dt when x is 4m from the building
form the image description of the problem, we see two-like triangles with the same base and height ratios
so
2 / (12-x) = y / 12
24 = y(12 - x )
y = 24 / (12-x)
dy/dt = 24/(12-x)² × dx/dt
Now at x = 4,
we substitute
dy/dt will be;
⇒ 24/(12 - 4)² × -2.3
= 24/64 - 2.3
= 0.375 × -2.3
dy/dt = - 0.8625 m/s
Therefore, the rate of the change of the length of the shadow is - 0.8625 m/s.
The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.
Answer:
The height will be 4 times.
Explanation:
Given that,
The speed at the bottom of the hill doubled.
We need to calculate the height
Using conservation of energy




Therefore,

Here, m and g are constant
Hence, The height will be 4 times.
Answer:
E = 10⁵ J
Explanation:
given,
Power, P = 100 TW
= 100 x 10¹² W
time, t = 1 ns
= 1 x 10⁻⁹ s
The energy of a single pulse is:-
Energy = Power x time
E = P t
E = 100 x 10¹² x 1 x 10⁻⁹
E = 10⁵ J
The energy contained in a single pulse is equal to 10⁵ J