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9966 [12]
3 years ago
12

PLZZZ HELP!!! State the forces in a spring, magnets, electric scales, and a newton metre

Physics
1 answer:
Lisa [10]3 years ago
5 0

Answer:

The various kinds of forces that exist in the items below are:

Spring = Spring Force

Magnet = Magnetic Force

Electric Scale = Electric Force

Newton Metre = Spring Force

Explanation:

Spring Force:

When you fix an object unto or upon a spring, the resulting compression or stretching exerted by that object on the spring is called the Spring Force. Conversely, the object is acted upon by the force that brings back the object to its state of equilibrium of rest.

Magnetic Force:

The motion between electrically charged particles always results in magnetic force. This is the primary force that is active in electric motors and normal magnets.

Electric Force

In order to create an electric force, you'd need to charge two particles and bring them in proximity to one another. Usually, they'll either repel or attract one another. This attraction or repulsion is known as an electric force.

Cheers

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Answer:

a_{cp}=7.77m/s^2

Explanation:

The equation for centripetal acceleration is a_{cp}=\frac{v^2}{r}.

We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.

Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference C=2\pi r and the period T, then we have:

v=\frac{C}{T}=2\pi r f

Putting all together:

a_{cp}=\frac{(2\pi r f)^2}{r}=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2

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3 years ago
What is the acceleration of a 4,000 kg car pushed with a<br> force of 12,000 N?
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Answer:

3 m/s

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3 years ago
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The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti
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If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the
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Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = -10 m

y_{0}: is the initial height = 0

v_{0_{y}}: is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

x_{f} = 5.51 m/s*1.43 s = 7.88 m

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

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3 years ago
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combining them we have:

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