Answer:
t = 5.19 s
Explanation:
We have,
Height of the cliff is 132 m
It is required to find the time taken by the ball to fall to the ground. Let t is the time taken. So, using equation of kinematics as :
So, it will take 5.19 seconds to fall to the ground.
False because opposites attract. :)
Answer: The height of the cloud = 394.55 m
Explanation:
The observer is 500m away from the spotlight.
Let x be the distance from the observer to the interception of the segment of the height, h with the floor. The equations are thus:
Tan 45° = h/x ... eq1
Tan 75° = h/(500- x ) ... eq2
From eq 1, Tan 45° = 1, therefore eq1 becomes:
h = x ... eq3
Put eq3 into eq2
Tan 75° = h/(500- h)
h = ( 500 - h ) Tan 75°
h = 500Tan 75° - hTan75°
h + h Tan 75° = 500 Tan 75°
h ( 1 + Tan 75° ) = 500 Tan75°
h = 500Tan75°/ (1 + Tan 75°)
h= 1866.02 / 4.73
h = 394.55m
Answer:
h = 18.41 m
Explanation:
Given that,
Mass of a test rocket, m = 11 kg
Its fuel gives it a kinetic energy of 1985 J by the time the rocket engine burns all of the fuel.
According to the law of conservation of energy,
PE = KE = mgh
h is height will the rocket rise
So, the rocket will rise to a height of 18.41 m.
Answer:
vT = v0/3
Explanation:
The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R
Since F = F0 = F'
GMm/R² = m(v0)²/R
GM = (v0)²R (1)
Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R
Since F1 = F'
GMm/27R² = m(vT)²/3R
GM = 27(vT)²R/3
GM = 9(vT)²R (2)
Equating (1) and (2),
(v0)²R = 9(vT)²R
dividing through by R, we have
9(vT)² = (v0)²
dividing through by 9, we have
(vT)² = (v0)²/9
taking square-root of both sides,
vT = v0/3
