The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
Answer:
F = 800 [N]
Explanation:
To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.
We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.
ΣPbefore = ΣPafter

where:
m₁ = mass of the hammer = 0.15 [m/s]
v₁ = velocity of the hammer = 8 [m/s]
F = force [N] (units of Newtons)
t = time = 0.0015 [s]
v₂ = velocity of the hammer after the impact = 0
![(0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N]](https://tex.z-dn.net/?f=%280.15%2A8%29-%28F%2A0.0015%29%20%3D%20%280.15%2A0%29%5C%5CF%2A0.0015%20%3D%200.15%2A8%5C%5CF%20%3D%201.2%2F%280.0015%29%5C%5CF%20%3D%20800%20%5BN%5D)
Note: The force is taken as negative since it is exerted by the nail on the hammer and this force is directed in the opposite direction to the movement of the hammer.
Answer: Their u go i found it their was about 3 pages i did not no what pages u had to do.
Explanation:
Answer:
A) T.
Explanation:
Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:
So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.
Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.
From this question, the plane is still up in the air.
We have wind blowing in [W 60° N ]
To solve the problem we have to make use of the sine rule

We put the values in the equation, we have:
50/Sinθ = 200/sin60°
The next step is to cross multiply
50 x sin60° = 200Sinθ
50 x 0.8660 = 200sinθ
We make Sin θ the subject
Sine θ = 43.30/200
sine θ = 0.2165
we find the value of θ
θ = sine⁻¹(0.2165)
θ = 12.50
So Lindsay has to fly this plane towards this direction
[W 12.5° S]
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