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vladimir1956 [14]
3 years ago
12

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

ngular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.7m. If the thrower takes 1.2s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Physics
1 answer:
harina [27]3 years ago
6 0

Answer:

4.437 m/s

Explanation:

Diameter of rotation d is 1.7 m

Radius of rotation = d/2 = 1.7/2 = 0.85 m

If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s

We convert to rad/s

Angular speed = 2 x pi x 0.83

= 2 x 3.142 x 0.83 = 5.22 rad/s

Speed is equal to the angular speed times the radius of rotation

Speed = 5.22 x 0.85 = 4.437 m/s

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However, this is missing an important element related to forces. We could expand it by stating:
An object at rest stays at rest and an object in motion stays in motion at a constant speed and direction unless acted upon by an unbalanced force.
By the time Newton came along, the prevailing theory of motion—formulated by Aristotle—was nearly two thousand years old. It stated that if an object is moving, some sort of force is required to keep it moving. Unless that moving thing is being pushed or pulled, it will simply slow down or stop. Right?
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Name one contact force and one force that acts through a force field
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A 2.50 kg ball moving at 7.50 m/s is caught by a 70.0 kg man while the man is standing on ice. What is the common velocity of th
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2 years ago
A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor
STatiana [176]

Answer:

The magnitude of the second charge is \rm 1.062\times 10^{-7}\ C or \rm 0.1062\ \mu C.

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge q_o at a point in the electric field of another charge q is given by the product of the amount of charge q_o and electric potential at that point due to the charge q.

U = q_o\ V.

The electric potential at that point is given by

V = \dfrac{kq}{r}.

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Therefore,

U=q_o\ \dfrac{kq}{r}.

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U_i = \dfrac{kq_1q_2}{\infty}=0.

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(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).

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r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.

The electric potential energy of the charges in this configuration is given by

U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

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