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dedylja [7]
3 years ago
12

I need help on question 5

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0
Weathering, erosion, impact, pressure  
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3 years ago
A 2.31 kg rope is stretched between supports 10.4 m apart. If one end of the rope is tweaked, how long will it take for the resu
zlopas [31]

Answer:

t = 0.657 s

Explanation:

First, let's use the appropiate equations to solve this:

V = √T/u

This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.

Where:

V: Speed of the disturbance

T: Tension of the rope

u: linear density of the rope.

The density of the rope can be calculated using the following expression:

u = M/L

Where:

M: mass of the rope

L: Length of the rope.

We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:

u = 2.31 / 10.4 = 0.222 kg/m

Now, replacing in the first equation:

V = √55.7/0.222 = √250.9

V = 15.84 m/s

Finally the time can be calculated with the following expression:

V = L/t ----> t = L/V

Replacing:

t = 10.4 / 15.84

t = 0.657 s

4 0
3 years ago
Which gas is responsible for polarizination defect?​
Alex_Xolod [135]

Answer:

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Read 2 more answers
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
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