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4 years ago

Since an object moving in uniform circular motion keeps a constant speed, there is no force necessary to keep it in motion.

Physics

2 answers:

e-lub [12.9K]4 years ago

8 0

Answer:

false

Explanation:As mentioned earlier in this lesson, an object moving in a circle is experiencing an acceleration. Even if moving around the perimeter of the circle with a constant speed, there is still a change in velocity and subsequently an acceleration. This acceleration is directed towards the center of the circle. And in accord with Newton's second law of motion, an object which experiences an acceleration must also be experiencing a net force. The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration. This is sometimes referred to as the centripetal force requirement. The word centripetal (not to be confused with the F-word centrifugal) means center seeking. For object's moving in circular motion, there is a net force acting towards the center which causes the object to seek the center.

To understand the importance of a centripetal force, it is important to have a sturdy understanding of the Newton's first law of motion - the law of inertia. The law of inertia states that ...

... objects in motion tend to stay in motion with the same speed and the same direction unless acted upon by an unbalanced force.

According to Newton's first law of motion, it is the natural tendency of all moving objects to continue in motion in the same direction that they are moving ... unless some form of unbalanced force acts upon the object to deviate its motion from its straight-line path. Moving objects will tend to naturally travel in straight lines; an unbalanced force is only required to cause it to turn. Thus, the presence of an unbalanced force is required for objects to move in circles.

ANEK [815]4 years ago

4 0

Answer:

False.

It would have been needed a push of force to start along with a way to keep it in motion.

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A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

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If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

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$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

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3 years ago

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Answer:

a. I = 0.76 A

b. Z = 150.74

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d. RL₂ = 602.58

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V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω

Using law of Ohm

V = I * R

I = Vc / Rc = 364 V / 473 Ω

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The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

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Z = 150.74

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

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RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -) √ 150.74² 77.0²

RL = 473 (+ / -) (129.58)

RL₁ = 34.41 , RL₂ = 602.58

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RL₂ = 602.58

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