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frutty [35]
4 years ago
10

A sign is held in equilbrium by 7 vertically hanging ropes attached to the ceiling. If each rope has an equal tension of 53 Newt

ons, what is the mass of the sign in kg?
Physics
1 answer:
stiks02 [169]4 years ago
5 0

Answer:

37.86 kg

Explanation:

The weight of sign board is equally divided on each rope. It means the tension in all the ropes is equal to the weight of the sign board in equilibrium condition.

Tension in each rope = 53 N

Tension in 7 ropes = 7 x 53 N = 371 N

Thus, The weight of sign = 371 N

Now, weight = m g

where m is the mass of sign.

m = 371 / 9.8 = 37.86 kg

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A beam of white light passes through a uniform thickness of air. If the intensity of the scattered light in the middle of the gr
k0ka [10]

This question is incomplete, the complete question is;

A beam of white light passes through a uniform thickness of air. If the intensity of the scattered light in the middle of the green part of the visible spectrum (λ=520nm) is I, find the intensity (in terms of I) of scattered light

a) In the middle of the red part of the spectrum (λ= 665 nm)

b) In the middle of the violet part of the spectrum (λ = 420 nm)

Answer:

a) the intensity of scattered light ( in terms of I ) In the middle of the red part of the spectrum is 0.3739I

b) the intensity of scattered light ( in terms of I ) In the middle of the violet part of the spectrum is 2.3497I  

Explanation:

Given the data in the question,

the visible spectrum (λ=520nm) = I

we know that; intensity of scattered light is proportional to 1 / λ⁴  

I ∝ ( 1 / λ⁴ )

so

a)

I_R / I = ( λ / λ_R )⁴

As the middle of the green part of the visible spectrum λ is 520nm and middle of the red part of the spectrum λ_R is 665 nm

we substitute

I_R / I = ( 520 / 665 )⁴

I_R / I = ( 0.781954887 )⁴

I_R / I = 0.3739

I_R  =  0.3739I  { in terms of I  }

Therefore, the intensity of scattered light ( in terms of I ) In the middle of the red part of the spectrum is 0.3739I

b)

I_V / I = ( λ / λ_V )⁴

As the middle of the green part of the visible spectrum λ is 520nm and middle of the red part of the spectrum λ_R is 420 nm

we substitute

I_V / I = ( 520 / 420 )⁴

I_V / I = ( 1.238095 )⁴

I_V / I = 2.3497

I_V = 2.3497I  { in terms of I  }

Therefore, the intensity of scattered light ( in terms of I ) In the middle of the violet part of the spectrum is 2.3497I  

5 0
3 years ago
Read 2 more answers
What are 5 examples of pseudoscience that can be found on the Internet?
mel-nik [20]

-- flat Earth
-- no moon landings
-- no gravity
-- chem trails
-- astrology

7 0
3 years ago
Commercials for a toy bouncy ball advertise that it will bounce to a height that is greater than the
storchak [24]

Answer:

because energy will be lost due to friction, sound, and heat (arguably similar to friction) and ENERGY MUST STAY THE SAME so it is IMPOSSIBLE for the ball to bounce higher than when dropped!

7 0
4 years ago
A 7.0kg skydiver is descending with a constant velocity
Vikentia [17]

Answer:

The air resistance on the skydiver is 68.6 N

Explanation:

When the skydiver is falling down, there are two forces acting on him:

- The force of gravity, of magnitude mg, in the downward direction (where m is the mass of the skydiver and g is the acceleration due to gravity)

- The air resistance, R, in the upward direction

So the net force on the skydiver is:

F=mg-R

where

m = 7.0 kg is the mass

g=9.8 m/s^2

According to Newton's second law of motion, the net force on a body is equal to the product between its mass and its acceleration (a):

F=ma

In this problem, however, the skydiver is moving with constant velocity, so his acceleration is zero:

a=0

Therefore the net force is zero:

F=0

And so, we have:

mg-R=0

And so we can find the magnitude of the air resistance, which is equal to the force of gravity:

R=mg=(7.0)(9.8)=68.6 N

6 0
4 years ago
A mass m at the end of a spring vibrates with a frequency
Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

F = -kx

And Newton's Second Law also states that

F = ma = m\frac{d^2x}{dt^2}

Combining two equations yields

a = -\frac{k}{m}x

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

\omega = \sqrt{\frac{k}{m}}

And given that ω = 2πf

the relation between frequency and mass becomes

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Let's apply this to the variables in the question.

0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg

6 0
3 years ago
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