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mafiozo [28]
3 years ago
11

What is N stands for?

Physics
1 answer:
Paladinen [302]3 years ago
8 0
Newton, a unit of force equal to the force that imparts an acceleration of 1 m/sec/sec to a mass of 1 kilogram; equal to 100,000 dynes
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What is the relationship between an object’s temperature and its heat?
Over [174]

Answer:

Explanation:

Heat and temperature are related to each other, but are different concepts. Heat is the total energy of molecular motion in a substance while temperature is a measure of the average energy of molecular motion in a substance. ... Temperature does not depend on the size or type of object

3 0
4 years ago
What force is applied to a 60 kg person if it takes 7.8 seconds to reach a speed of 12.86 m/s from rest?
docker41 [41]

Answer:

drag

Explanation:

drag is the force that acts against any moving object.

if it takes him any amount of time to get from one speed to another, it is caused by drag.

3 0
2 years ago
Question 26 Unsaved
vlabodo [156]
The density increases
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3 years ago
What is the relationship between mass and inertia? Give an example
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Mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it ha

5 0
3 years ago
When mass m is tied to the bottom of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 180 h
Katena32 [7]
The frequency of the nth-harmonic of a string is given by
f_n =  \frac{n}{2L}  \sqrt{ \frac{T}{\mu} }
where n is the number of the harmonic, L is the length of the string, T the tension and \mu the linear density. 

In our problem, since the mass m is tied to the string, the tension is equal to the weight of the object tied:
T=mg
Substituting into the first formula, we have
f_n =  \frac{n}{2L}  \sqrt{ \frac{mg}{\mu} }

In our problem we have n=2 (second harmonic). In the previous equation, the only factor which is not constant between the first and the second part of the problem is m, the mass. So, we can rewrite everything as
f_2 = K  \sqrt{m}
where we called 
K= \frac{2}{2L}  \sqrt{ \frac{g}{\mu} }

In the first part of the problem, the mass of the object is m and f_2 = 180 Hz. So we can write 
180 Hz = K  \sqrt{m}

When the mass is increased with an additional 1.2 kg, the relationship becomes
270 Hz = K \sqrt{(m+1.2 Kg)}

By writing K in terms of m in the first equation, and subsituting into the second one, we get
180 Hz  \sqrt{ \frac{m+1.2 Kg}{m} }=270 Hz
and solving this, we find
m=0.96 kg


5 0
4 years ago
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