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Mrrafil [7]
3 years ago
13

The rules for the shot put specify that the circle from which the shot-putter puts the shot must be only 7 ft 6 in. in diameter.

The athlete may not allow any part of his or her body to touch anything outside of the circle during the throw. If the throwing circle were enlarged to be 10.0 ft in diameter, would performances in the shot put improve or decline? Why?
Physics
2 answers:
notsponge [240]3 years ago
7 0

Answer:

His performance would improve

Explanation:

This is because, he now has a larger distance to cover in the 10.0 ft diameter circle. This increased distance would cause him to give a greater angular speed to his whirl and thus a greater tangential velocity to the shot put. This greater speed would thus cause the shot put to travel more in the 10.0 ft diameter circle throw than in the 7 ft 6 in diameter circle throw.

ss7ja [257]3 years ago
5 0

Answer:

The performance will improve because more energy will be expended on the shot

Explanation:

The performance in the shot put is the work done or the energy expended by the short putter

Work done = Energy expended = Force applied by the short putter * displacement

When the diameter of the shot = 7ft 6 in = 7.5 ft

Work done = 7.5 * Force

When the diameter of the shot = 10 ft

Work done = 10 * force

Since the circle is enlarged, shot putter will be able to do more work on the shot, causing an increase in the expended energy, hence an improvement in performance

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Note that the simulation allows you to also display the force of the smaller moon
Lelu [443]

the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet

Explanation:

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F=G\frac{m_1 m_2}{r^2}

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4 0
3 years ago
A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially
Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

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==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t   ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt=  6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

4 0
3 years ago
Calculate the force needed to bring a 1,019-kg car to rest from a speed of 29 m/s in a distance of 118 m
son4ous [18]
Acceleration = ▵v/▵t
Time = d/v
Fisrt calculate time : ( 118/29 ) = 4 seconds
Then calculate acceleration
A = 29/4 = 7.25 m/s²
Now the force.
Force = mass * acceleration.
F= 1,019 * 7.25
F= 7,387 N
8 0
3 years ago
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