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salantis [7]
2 years ago
12

if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *

0.013 n 4 n 0. 32 n 20.1 n
Physics
1 answer:
SSSSS [86.1K]2 years ago
4 0

Answer:

0.013N

Explanation:

F=\frac{mv^{2} }{r}

m=0.04, v=4.4m/s, r=60

F=0.013 N

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Newton’s second law states that the acceleration of an object is dependent on 1 variable which is the net force
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I think it’s false. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and then mass of the object.
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3 years ago
Which of the following statements apply to electric charges?
Gre4nikov [31]

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

6 0
3 years ago
Where can radiation be found in nature and how is it affected
Murljashka [212]
The sun is a clear example of objects releasing radiation in nature
8 0
2 years ago
Which of the following circuits will stop producing light if one bulb burns out? *
gulaghasi [49]

Answer:

Circuit 1 and Circuit 3

Explanation:

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6 0
3 years ago
You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b
Radda [10]

Answer:

\theta=34 \textdegree

Explanation:

From the question we are told that:

Mass m=55kg

Angle \theta =28.0

Coefficient of static friction \alpha =0.680

Generally, the equation for Newtons second Law is mathematically given by

For

\sum_y=0

N=mgcos \theta

for

\sum_x=0

F_{s}=mgsin\theta

Where

F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta

F_{s}=0.68*55*9.8*cos 28

F_{s}=323.62N

Therefore

\alpha mgcos \theta=mg sin \theta

\theta=tan^{-1}(0.68)

\theta=34 \textdegree

6 0
2 years ago
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