The forklift does no work on the box at all.
And work doesn't have a direction.
Ok, so you've got to figure out a force F and you have the speed in which the boxer punches on determinate time and the mass of the sheet of paper.
So based on the formula that says that the Force is equal to the mass multiplied by the acceleration => F=ma.
You look at it and see that you only have mass which is measured on KG so there is no problem.
then you have the acceleration which is measured on meters and is defined by: a = Δv/Δt
So now you can replace the velocity and the time you have there
⇒ a 25m/s / 0.05s
you have computing that ⇒ 50m because the seconds were cancelled out.
and then you plug the meters into the force equation.
F=(0.005kg)(50)
F=0.25N
so the boxer will have a force of 0.25 Newton's.
Answer:
(C) an increase in tue distance between the ibject causes a greater change in the gravitational force than the same increase in mass
Hope this helps
The rate in witch ditermans the speed or vibration of the movment under the waves witch couses vibrational freequencys to be disrupted.
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
