<em>Steel: 11.0 – 12.5</em>
<em>T̶e̶t̶s̶u̶t̶e̶t̶s̶u̶ ̶T̶e̶t̶s̶u̶t̶e̶t̶s̶u̶</em>
Thanks,
<em>Deku ❤</em>
Answer:
Explanation:
Let the velocity be v
Total energy at the bottom
= rotational + linear kinetic energy
= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell = mr² )
= 1/2 mr²ω² + 1/2 mv² ( v = ω r )
= 1/2 mv² +1/2 mv²
= mv²
mv² = mgh ( conservation of energy )
v² = gh
v = √gh
= √9.8 x 1.8
= 4.2 m /s
-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s
-- The acceleration of gravity is 9.8 m/s².
-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.
-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================
-- The horizontal component of the ball's velocity is 14 cos(</span><span>51°) = 8.81 m/s
-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.
As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
Answer:
Power_input = 85.71 [W]
Explanation:
To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

where:
W = work [J] (units of Joules)
F = force [N] (units of Newton)
d = distance [m]
We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.
Now replacing:
![W = (80*10)*3\\W = 2400 [J]](https://tex.z-dn.net/?f=W%20%3D%20%2880%2A10%29%2A3%5C%5CW%20%3D%202400%20%5BJ%5D)
Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

where:
P = power [W] (units of watts)
W = work [J]
t = time = 40 [s]
![P = 2400/40\\P = 60 [W]](https://tex.z-dn.net/?f=P%20%3D%202400%2F40%5C%5CP%20%3D%2060%20%5BW%5D)
The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.
![Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]](https://tex.z-dn.net/?f=Effic%3D0.7%5C%5CEffic%3DP_%7Brequired%7D%2FP_%7Bintroduced%7D%5C%5CP_%7Bintroduced%7D%3D60%2F0.7%5C%5CP_%7Bintroduced%7D%3D85.71%5BW%5D)
<h2>
Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>
Explanation:
Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.
That is force 1 is 100 pound with x axis at 20°
F₁ = 100 cos 20 i + 100 sin 20 j
F₁ = 93.97 i + 34.20 j
That is force 2 is 150 pound with x axis at 60°
F₂ = 150 cos 60 i + 150 sin 60 j
F₂ = 75 i + 129.90 j
F₁ + F₂ = 93.97 i + 34.20 j + 75 i + 129.90 j
F₁ + F₂ = 168.97 i + 164.10 j

Resultant is 235.54 pounds at an angle 44.16° to X axis.