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andrew-mc [135]
3 years ago
11

Ac=v^2/r solve for v, solve for r

Physics
2 answers:
bazaltina [42]3 years ago
8 0

Answer:

v = √ac×r

r = \frac{v^2}{ac}

Rasek [7]3 years ago
8 0

Answer:

Explanation:

Sorry cant help

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6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t
Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of  12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

y = u sin \theta - \dfrac{1}{2}gt^2................(2)

from equation(1) and (2)

y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}..........{3}

12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}

\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29

\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2

u = \sqrt{10959.34}

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0
3 years ago
How can we balance the use of fossil fuels with other forms of energy?
KiRa [710]
We can use renewable sources
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3 years ago
According to the graph of displacement vs. time, what is the object's displacement at time = 60 s?
krek1111 [17]

Answer:

The straight line that is obtained, intercept it on the y-axis and the value of displacement will obtained.

Explanation:

4 0
3 years ago
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In the pulley system shown below, a 360 N weight is slowly lifted. Assuming the system is 100% efficient and each pulley is weig
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Your answer is D. 361 N
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3 years ago
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A 110 kg quarterback is running the ball downfield at 4.5 m/s in the positive direction when he is tackled head-on by a 150 kg l
melisa1 [442]

Answer:

v_f=-0.29\frac{m}{s}

Explanation:

The principle of conservation of momentum, states that if the sum of the forces acting on a system is null, the initial total momentum of the system before a collision equals the final total momentum of the system after the collision. The collision is completely inelastic, which means that the players remain stick to each other after the collision:

p_i=p_f\\m_1v_1+m_2v_2=(m_1+m_2)v_f\\v_f=\frac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\v_f=\frac{(110kg)4.5\frac{m}{s}+150kg(-3.8\frac{m}{s})}{(110kg+150kg)}\\v_f=-0.29\frac{m}{s}

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3 years ago
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