Both options 5 and 6
Explanation:
Let us consider option 5,
In option 5 body is moving up with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.
Let us consider option 6,
In option 6 body is moving down with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.
Answer:
a) = 258352.5J
b) = 23.63 m/s
c) = 1.8m
Explanation:
Data;
Mass = 925kg
Distance (s) = 28.5m
Force constant (k) = 8.0*10⁴ N/m
g = 9.8 m/s²
a) = work = force * distance
But force = mass * acceleration
Force = 925 * 9.8 = 9065N
Work = F * s = 9065 * 28.5 = 258352.5J
b) acceleration (a) = (v² - u²) / 2s
a = v² / 2s
v² = a * 2s
v² = 9.8 * (2 * 28.5)
v² = 9.8 * 57
v² = 558.6
v = √(558.6)
V = 23.63 m/s
C). The work stops when the work done to raise the spring equals the work done to stop it by the spring
W = ½kx²
258352.5 = ½ * 8.0*10⁴ * x²
(2 * 258352.5) = 8.0*10⁴x²
516705 = 8.0*10⁴x²
X² = 516705 / 8.0*10⁴
X² = 6.46
X = √(6.46)
X = 2.54m
The compression was about 2.54m
Answer:
The force becomes 16 times what it is now.
Explanation:
The formula for gravitational force is
F = G * m1 * m2 / r^2
When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.
If you double m1 and m2 and don't do anything to r, you've already increased the force by 4 times. (2m1 * 2m2 = 4 * m1 * m2)
But you are not finished. If you 1/2 the distance, you are again increasing the Force by 4 times. 1 / (2r) ^2 = 1/ 4* r^2
Because this is in the denominator, the 1/4 is going to flip to the numerator.
So the total increase is going to be 4 * (4 * m1 * m2) = 16 * m1 * m2.
Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:
Answer:
h=2.86m
Explanation:
In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,
There is no kinetic energy in the initial state, nor potential energy in the end,
In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,
The inertia of the bodies is given by the equation,
On the other hand the angular velocity is given by
Replacing these values in the equation,
Solving for h,
