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3 years ago

What is the direction of the net force that acts on an object undergoing uniform circular motion?

Physics

2 answers:

krok68 [10]3 years ago

7 0

Answer: Option (D) is the correct answer.

Explanation:

An object moves with a constant speed when it has uniform circular motion. Due to change in direction this object accelerates.

Therefore, net force acts towards the center of circle which is also said to be inward force or centripetal force.

Thus, we can conclude that direction of the force is toward the center of the object's circular path that acts on an object undergoing uniform circular motion.

8_murik_8 [283]3 years ago

5 0

D. The direction of the force is toward the center of the object's circular path.

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You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?

In-s [12.5K]

-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is 14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.

As usual when we're discussing this stuff, we completely ignore air resistance.
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4 0

3 years ago

Read 2 more answers

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

3 years ago

Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg

Yanka [14]

Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.

6 0

4 years ago

Read 2 more answers

Please answer correctly to be marked as brainly.

lidiya [134]

A- PRICE

B-QUANTITY

C-SUPPLY

D-DEMAND

E-EQUILIBRIUM POINT

Explanation:

It is the Supply Demand curve in Economics. It gives relationship between price and quantity

7 0

3 years ago

Banked Curves: A 600-kg car is going around a banked curve with a radius of 110 m at a steady speed of 24.5 m/s. What is the app

Aleks04 [339]

The angle of baking from the calculation is obtained as 30°.

<h3>What is banking?</h3>

The term banking refers to a means of preventing vehicles from skidding off the road at curves.

We know that the banking angle is obtained from;

θ = tan-1(v^2/rg)

v = 24.5 m/s

r = 110 m

g = 9.8 m/s^2

θ = tan-1(25^2/9.8 * 110)

θ = tan-1(625 /1078)

θ = 30°

Learn more about the banking angle:brainly.com/question/26759099?r

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8 0

2 years ago