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WITCHER [35]
2 years ago
11

an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

nd. The radius of the hammer's path is 2.0 meters.
Physics
1 answer:
Semenov [28]2 years ago
6 0

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

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

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

<u>Fc = 504 N</u>

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3 years ago
A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of
PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box.  There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

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3 years ago
You are watching a magic show. For one trick the magician rolls a ball down a hill. Suddenly the ball stops moving down the hill
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8 0
3 years ago
Read 2 more answers
Find an equation in x and y for the line tangent to the curve ()=4,()=cos() at the point where =4.
m_a_m_a [10]

An equation in x and y for the line tangent to the curve ()=4,()=cos() at the point where =4 is x(t)=2t+2,y(t)=t^4.

<h3>What is tangent?</h3>

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7 0
1 year ago
Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di
Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, u=20m/s

And the mass of the car is, m=1000 kg

The total distance covered by the car before stop, s=45m

And the final speed of the car is, u=0m/s

Now initial kinetic energy is,

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Substitute the value of u and m in the above equation, we get

KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J

Now final kinetic energy is,

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Substitute the value of v and m in the above equation, we get

KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J

Now applying work energy theorem.

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