acceleration of rocket is given here as
now we know that
now integrating both sides
here since its given that rocket will accelerate for t = 10 s
so here we have
so after t = 10 s the speed of rocket will be 130 m/s upwards
<h2>Answer:</h2>
<u>This term shows the </u><u>mass of the space shuttle</u>
<h2>Explanation:</h2>
We know that the mass of the Earth is 5.972 × 10^24 kg. Similarly the sum of mass of earth and the mass of shuttle must be a greater number as compared to the number given. It simply means that the mass of earth is itself 5.972 × 10^24 kg and the value given is 3 × 105 kg so it is obvious that if was the sum then it must be greater than the mass of earth. Therefore we can say that this not the mass of earth, neither the sum of mass of earth and shuttle, but this is only the mass of space shuttle which is the last multiple choice.
Answer:
option B
Explanation:
given,
Force exerted by the hydraulic jack piston = F₁ = 250 N
diameter of piston, d₁ = 0.02 m
r₁ = 0.01 m
diameter of second piston, d₂ = 0.15 m
r₂ = 0.075 m
mass of the jack to lift = ?
now,
F₂ = 14062.5 N
F = m g
m = 1435 Kg
hence, the correct answer is option B
The position function x(t) of a particle moving along an x axis is 
a) The point at which particle stop, it's velocity = 0 m/s
So dx/dt = 0
0 = 0- 12t = -12t
So when time t= 0, velocity = 0 m/s
So the particle is starting from rest.
At t = 0 the particle is (momentarily) stop
b) When t = 0
SO at x = 4m the particle is (momentarily) stop
c) We have
At origin x = 0
Substituting
t = 0.816 seconds or t = - 0.816 seconds
So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
