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givi [52]
3 years ago
8

Transverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What i

s the mass of the string

Physics
1 answer:
Kipish [7]3 years ago
3 0

Answer:

0.055 kg

Explanation:

Given that

Length of the string, l = 5 m

Speed of the wave, v = 30 m/s

Tension on the string, F(t) = 10N

From the formula written in the attachment, we have

v = velocity of the wave, in m/s

F(t) = Tension on the string, in N

U = Mass per length of the string, in kg/m

m = Mass of the string, in kg

l = Length of the string, in m

See attachment for the calculation

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Trial 1: Get a textbook and put a sheet of paper on top of it. Fold the paper as needed to keep the paper from sticking over the
siniylev [52]

Answer:

1)  the two objects reach the floor at the same time.

2)the book reaches the floor much earlier than the foil

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more

Explanation:

This interesting experiment has the following results

1) first case. Sheet on top of book

In this case the two objects reach the floor at the same time.

This shows that the acceleration in the two objects is the same and we call it the acceleration of gravity.

The speed of the body increases as it goes down linearly.

This occurs because the book that receives air resistance is much heavier, so the resistance has almost no effect on its movement, the sheet does not have the air resistance because it goes down next to the book.

2) second case. Book and sheet next to each other.

In this case the book reaches the floor much earlier than the foil.

This is because the resisting force of the air has almost no effect on the book and its movement is little affected by this force.

In the case of the blade, it has very little weight, therefore as its speed increases, the resistance force of the air rapidly equals the weight of the blade.

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so after this, since the acceleration is zero, it goes down at constant speed, this speed is called the terminal velocity.

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more.

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3 years ago
Explain why radiation is dangerous for humans
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Answer:

because it affects the attom in living things

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Twins Jody and Taylor are rearranging the furniture in their bedroom and want to move a dresser across the room. The dresser has
xeze [42]

Answer:

yes, They will be able to move the dresser.

Explanation:

sliding force 90N

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3 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

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3 years ago
The metal gold crystallizes in a face centered cubic unit cell with one atom per lattice point. When X-rays with λ = 1.436 Å are
Umnica [9.8K]

Answer:

 r =  1.45 Å

Explanation:

given,

λ = 1.436 Å

θ = 20.62°

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Radius of the gold atom = ?

using Bragg's Law

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the radius of a(n) gold atom. is equal to 1.45 Å

7 0
3 years ago
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