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dangina [55]
3 years ago
12

An alumium rod when measured with a steel scale both being at 25 degree celcius appears to be 1 meter long. If the scale is corr

ect at 0 degree celcius, what is the true length of the rod at 25 degree celcius? What wull be the length of rod at 0 degree celcius?​
Physics
2 answers:
natulia [17]3 years ago
5 0
Given:
temperature at which the rod is measured= 25°
length of aluminium rod measured at 25°C= 1 meter
temperature at which the scale is correct= 0°C
Linear expansivity of steel= 12* 10^-6
linear expansivity of aluminum= 26* 10^-6
To find:
True length of rod at 25°C
Length of rod at 0°C
Solution:
Let the length of the aluminum and steel rod at 0°C be L0A and L0S respectively.
Length of the steel scale at 25°C , thermal expansion is:
LA25 = LA0(1+ (αS - αA)25)
LA0 = LA25/ (1+ (αS-αA)t)
LA0 = 100/ (1+ (12*10^-6 - 26*10^-6)25)
LA0 ≈ 100.05 cm at 0°C
2. Let the length at 25°C of the aluminium trod be LA25
The actual measure of aluminum rod at 25°C, will be
LA25 = 100 * (1+ 26* 10^-6*25)
LA25 ≈ 100.06 cm
marin [14]3 years ago
3 0

Answer:

An almunium rod when measured with a steel scale,both being at 25°C appears to be 1meter long. If the scale is correct at 0°C

Explanation:

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Calculation:

Step-1:

It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.

The gauge pressure at a particular depth of ocean water is calculated as:

$$P=\rho g h$$

Here \rho is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.

Step-2:

Now we are substituting the values to calculate the pressure at the depth of 2452-m.

$$\\\begin{aligned}\\P&=\rho gh\\&=1030 (\text{ kg/m}^3)\times 9.8 (\text{ m/s}^2)\times 2452 \text{ m}\\&=24.75\times 10^6 \text{ Pa}\times\frac{1 \text{ atm}}{10.1325 \times10^4 \text{ Pa}}\\&=224.268 \text{ atm}\\\end{aligned}\\$$

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Calculate the frequency (v) in Hertz (Hz) of a wave that has a wavelength of 4.71 meters.
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Explanation:

Given parameters:

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Unknown:

Frequency  = ?

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3 years ago
Suppose that the loudspeaker in the problem had a mass of 500 kg and the ropes hung 20∘ from the vertical. Into which of the fol
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The following intervals would you expect T(tension) to fall : <u>2000 to 4000 N </u>

<h3>Further explanation</h3>

The force acting on a system with static equilibrium is 0

\large{\boxed{\bold{\sum F=0}}

(forces acting as translational motion only, not including rotational forces)

\displaystyle \sum F_x = 0\\\\ \sum F_y = 0

We complete the questions:

A 20-kg loudspeaker is suspended 2.0 m below the ceiling by two ropes that are each 30? from vertical.

Find the value of T, the magnitude of the tension in either of the ropes.

Express your answer in newtons.

Suppose that the loudspeaker in the problem has a mass of 500 kg and the ropes hung 20? from the vertical. Into which of the following intervals would you expect T to fall? You don't have to calculate Texactly to answer this question; just make an estimate.

500 to 1000 N

1000 to 2000 N

2000 to 4000 N

4000 to 8000 N

8000 to 16,000 N

In a 500-kg loudspeaker system and two ropes that are each 20° from vertical, we see the forces acting on the y axis (vertical)

\displaystyle \sum F_y = 0\\\\T1_y+T2_y-w=0\\\\T1~cos~20^o+T2~cos~20^o=m\times g

we assume g = 10 m/s², then :

\displaystyle 2T~cos~20=500\times 10\\\\T~cos~20=2500\\\\\boxed{\bold{T=26592 N}}}

So the value of T is between 2000 and 4000 N

<h3>Learn more</h3>

Newton's Law

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Keywords : the loudspeaker, ropes, Tension, mass, intervals

#LearnWithBrainly

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