Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 = × 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s
Let the key is free falling, therefore from equation of motion
.
Take initial velocity, u=0, so
.
As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula
From above substituting t,
.
Now substituting all the given values and g = 9.8 m/s^2, we get
.
Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.
Answer:
The time is
The speed is
Explanation:
From the question we are told that
The height of the cliff is
Generally from kinematic equation we have that
before the jump the persons initial velocity is u = 0 m/s
So
=>
Generally from kinematic equation
=>
=>