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Natali5045456 [20]
1 year ago
13

similar to the case in the experiment you use two rubber bands, side to side first, to hold a weight of 3 reference objects. the

n, as in the experiment, the bands are pulled apart in their free ends until they form an angle of 30 degrees with respect to the horizontal. what are the deformations observed in these bands? ...
Physics
1 answer:
Sever21 [200]1 year ago
7 0

Three units deformations observed in these bands.

<h3>What forces do a rubber band encounter?</h3>

Elastic force is the force that permits some materials to regain their former shape after being stretched or crushed. Thus, a stretched rubber band is subject to an elastic force.

The rubber band experiment uses a straightforward rubber band to show entropic force and a refrigeration cycle. The rubber band experiment involves stretching and then releasing a rubber band while measuring its temperature.

Always acting in the opposite direction of motion is friction. This indicates that if friction is there, it cancels out some of the force driving the motion (if the object is being accelerated). This results in a decreased acceleration and a smaller net force.

learn more about Elastic force refer

brainly.com/question/5055063

#SPJ14

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Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
A stone is thrown vertically upwards with an initial velocity of 20m/sec. Find the maximum height ot reaches and the time taken
MAXImum [283]

Answer:

The height reached is 20m, The time taken to reach 20m is 2 seconds

Explanation:

Observing the equations of motion we can see that the following equation will be most helpful for this question.

v^{2} = u^{2} + 2as

We are given initial velocity, u

We know that the stone will stop at its maximum height, so final velocity, v

Acceleration, a

And we are looking for the displacement (height reached), s

Substitute the values we are given into the equation

0^{2} = 20^{2} + 2(10)s

Rearrange for s

0^{2} -20^{2} =20s

-400=20s

\frac{-400}{20} =s

s = -20 (The negative is just showing direction, it can be ignored for now)

The height reached is 20m

Use a different equation to find the time taken

s = vt - \frac{1}{2} at^{2}

Substitute in the values we have

-20=(0)t - \frac{1}{2} (10)t^{2}

Rearrange for t

-20 =0 -5 t^{2}

\frac{-20}{-5} =t^{2}

4 = t^{2}

t = 2s

The time taken to reach 20m is 2 seconds

4 0
3 years ago
Newtown third law applies to blank of objects​
fomenos

Answer:

All

Explanation:

I'm not sure what you meant but Newton's third law which basically states that every action has an equal and opposite reaction applies to <em>all</em> objects. So I think the answer is all.

8 0
3 years ago
A 46.5-kg ball has a momentum of 57.2 kg m/s. What is the ball's speed?
Serhud [2]

Answer:

1.23 m/s

Explanation:

p=mv

57.2 = 46.5v

v= 57.2/46.5

v= 1.23

If you want to verify your answer, just insert the value of v in the equation.

5 0
2 years ago
5. This break-dancer's speed is not changing as he spins on his head, but he is
n200080 [17]

Answer:

the velocity is changing therefore the acceleration is changing too.

Explanation:

8 0
3 years ago
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