Answer:
120 km/hr
Explanation:
Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.
Now, at any instant, D² = d² + L²
= d² + 4²
= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.
differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have
d(D²)/dt = d(d² + 16)/dt
2DdD/dt = 2d[d(d)/dt]
dD/dt = 2d[d(d)/dt] ÷ 2D
Now d(d)/dt = vertical speed of rocket = 200 km/hr
dD/dt = 200d/D [D = √(d² + 16)]
dD/dt = 200d/[√d² + 16]
Now substituting d = 3 km, the distance the rocket has risen into the equation, we have
dD/dt = 200(3)/[√(3² + 16)]
dD/dt = 600/[√(9 + 16)]
dD/dt = 600/√25
dD/dt = 600/5
dD/dt = 120 km/hr
So, the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.
Answer:
Explanation:
Given that,
Initial speed, u = 5.1 m/s
Final speed, v = 6.2 m/s
Time, t = 25 s
We need to find acceleration. We know that,
acceleration = rate of change of velocity
So, her acceleration is .
To show stuff that we cant see in very well
Answer:
hello your question is incomplete below is the complete question
A 2.0 kg block starts from rest on a positive x axis 3.0m from the origin and thereafter has an acceleration given by a= 4.0i - 3.0jin m/s2. At the end of 2.0 s, its angular momentum about the origin is ______.
answer : L = ( 2 kg ) [(-18m^2/s)k]
Explanation:
velocity vector(v) = (4.0i - 3.0j )(2.0s)
position vector(r) = ( 3.0m)i
determine the angular momentum about the origin
L = m ( r * v )
attached below is the detailed solution