height = .0970 mm = .00831 inches
Volume = length * breadth * height = 8.5 * 11 * 0.00381
Volume = 0.776985 inches^cube
Answer:

Explanation:
We can use the following kinematics equations to solve this problem:
.
Using the first one to solve for acceleration:
.
Now we can use the second equation to solve for the distance travelled by the airplane:
(three significant figures).
Answer:
787528.7 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).
From the question,
W = Tcos∅(d)............. Equation 1
Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.
But,
d = v(t)..................... equation 2
Where v = velocity, t = time
Substitute equation 2 into equation 1
W = Tcos∅(vt)............. Equation 3
Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s
Substitute into equation 3
W = 210(cos23°)(1.94×2100)
W = 787528.7 J
Answer:
The work done by the weightlifter, W = 700 J
The power of the weightlifter, P = 350 watts
Explanation:
A weightlifter lifts a set of weights a vertical distance, s = 2 m
The force exerted to lift the weight, F = 350 N
The work done by the body is defined as the product of the force applied by the body to the displacement it caused.
W = F x s
= 350 N x 2 m
= 700 J
The work done by the weightlifter, W = 700 J
The time taken by the weightlifter to lift the weight, t = 2 s
The power is defined as the rate of body to do work. It is given by the equation,
P = W / t
= 700 J / 2 s
= 350 watts
Hence, the power of the weightlifter, P = 350 watts