An Austin volleyball player bumps a 5 kg ball into the air. It reaches a height of 2.8 meters. How fast was the ball going as it
1 answer:
Answer:
v = 7.4 m/s
Explanation:
Given that,
Mass if a volleyball, m = 5 kg
The ball reaches a height of 2.8 m
We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :
So, the required speed is 7.4 m/s. Hence, the correct option is (b).
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2.50 miles is equal to 4026 m.
<u>Explanation:</u>
As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.
As 1 km = 0.621 miles, then
So, 2.50 miles will be equal to
Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km. So,
Thereby,
Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
