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Vlad1618 [11]
3 years ago
15

which of the following are examples of circular motion? A. Roller skating down a hill. B. A race car going around a rounded curv

e in the road. C. Earth going around the sun. D. The path of a soccer ball rolling on the ground. E. A basketball falling toward the hoop after being thrown.
Physics
1 answer:
Katena32 [7]3 years ago
6 0

Answer:

C. Earth going around the sun.

Explanation:

Circular motion should have a center to repeat its motion.

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One of the explorers is able to jump a maximum distance of 19.5 m with an initial speed of 2.80 m/s. Find the gravitational acce
BlackZzzverrR [31]

Answer:

0.2 m/s^2

Explanation:

Maximum distance, h = 19.5 m

initial speed, u = 2.80 m/s

final speed, v = 0 m/s

Let the gravitational acceleration is a.

Use third equation of motion

v^{2}=u^{2}-2\times a\times h

0^{2}=2.80^{2}-2\times a\19.5

2\times a\19.5=7.84

a = 0.2 m/s^2

Thus the gravitational acceleration on the exoplanet is 0.2 m/s^2

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3 years ago
A=vf-vi/t is the equation for calculating the acceleration of an object. write out the relationship shown in the equation using
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Basically it is the difference in velocity divided by the time it takes to make that change.
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4 years ago
A.to reduce<br> b.to dispose<br> c.to prevent<br> d.to help
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Answer:

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5 0
3 years ago
You have a spool of copper wire 4.82 mm in diameter and a power supply. You decide to wrap the wire tightly around a soda can th
GalinKa [24]

To solve the problem it is necessary to have the concepts of the magnetic field in a toroid.

A magnetic field is a vector field that describes the magnetic influence of electric charges in relative motion and magnetized materials.

By definition the magnetic field is given by the equation,

B=\frac{\mu_0 NI}{2\pi r}

Where,

\mu_0 = Permeability constant

N = Number of loops

I = Current

r = Radius

According to the given data we have that the length is 120mm and the thickness of the copper wire is 4.82mm.

In this way the number of turns N would be

N=\frac{120mm}{4.82mm}

N = 24.89 \approx 25 turns

On the other hand to find the internal radius, we know that:

2\pi r_i = 12cm

r_i= \frac{12}{2\pi}

r_i= 1.91cm

Therefore the total diameter of the soda would be

r= r_i+r_o = 1.91+6.5=8.51cm

Applying the concept related to magnetic field you have to for the internal part:

B_i=\frac{\mu_0 NI}{2\pi r_i}

B_i=\frac{(4\pi*10^{-7}) (25)(230)}{2\pi (1.91*10^{-2})}

B_i = 0.060T

The smallest magnetic field would be on the outside given by,

B_o=\frac{\mu_0 NI}{2\pi r}

B_o=\frac{(4\pi*10^{-7}) (25)(230)}{2\pi 8.51}

B_o = 0.0136T

<em>Therefore the maximum magnetic field is 0.06T.</em>

4 0
4 years ago
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