Answer:
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Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
Answer: 31.6ft
Explanation:
Check the attachment for the diagram.
According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|
that is 20ft - 10ft = 10ft
According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.
|AE|^2 + |EC|^2 = |AC|^2
10^2 + 30^2 = |AC|^2
100 + 900 = |AC|^2
|AC| = √1000
|AC| = 31.6ft
Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.
Answer:
A. W = 6875.0 J.
B. W = -14264.6 J.
Explanation:
A. The work done by the rider can be calculated by using the following equation:

Where:
: is the force done by the rider = 25 N
d: is the distance = 275 m
θ: is the angle between the applied force and the distance
Since the applied force is in the same direction of the motion, the angle is zero.

Hence, the rider does a work of 6875.0 J on the bike.
B. The work done by the force of gravity on the bike is the following:
The force of gravity is given by the weight of the bike.
And the angle between the force of gravity and the direction of motion is 180°.
The minus sign is because the force of gravity is in the opposite direction to the motion direction.
Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.
I hope it helps you!
Answer:
m = 2.01[kg]
Explanation:
This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

where:
F = force = 12.5 [N]
m = mass [kg]
a = acceleration = 6.2 [m/s²]
![12.5=m*6.2\\m = 2.01[kg]](https://tex.z-dn.net/?f=12.5%3Dm%2A6.2%5C%5Cm%20%3D%202.01%5Bkg%5D)