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Stolb23 [73]
3 years ago
7

30. The length of mercury thread when it is at 0°C, 100°C and at an unknown temperature 0 is 25mm, 225mm and 175mm respectively.

What is the value of 02 A. 85.0°C B. 80.0°C C. 75.0°C D. 70.0°C ​
Physics
1 answer:
yulyashka [42]3 years ago
3 0

Answer:

75°C

Explanation:

175−25/225−25=x−0/100−0

150/200=x/100

x=150×100/200

= 75°C

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A particle is moving with SHM of period pie . initially it is 10 cm from The center of the motion and moving in the positive dir
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Answer:

y = 10.44cos(2t - 0.291) cm

Explanation:

y = Acos(2πt/T + φ) = Acos(2πt/π + φ) = Acos(2t + φ)

v = y' = -2Αsin(2t + φ)

10 = Acos(2(0) + φ) = Acosφ

6 = -2Αsin(2(0) + φ) = -2Asinφ

6/10 = -2Asinφ/Acosφ = -2tanφ

tanφ = -0.3

φ = -0.291 radians

10 = Acos(-0.291)

A = 10/cos(-0.291) = 10.44

7 0
3 years ago
A speed does not involve the element of
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A speed does not involve the element of direction.
6 0
3 years ago
If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
kaheart [24]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

5 0
3 years ago
Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
Tanzania [10]
If you increase the mass m of the car, the force F will increase, while acceleration a is kept constant. Because F and m are directly proportional.
If you increase the acceleration a of the car, the force F will increase, while mass m is kept constant. Because F and a are directly proportional.

How can Newton's laws be verified experimentally; is by setting this experiment, and changing one variable while keeping the other constant, then observe the change in F.

Hope this helps.
5 0
3 years ago
Read 2 more answers
An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet
Harrizon [31]

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

 

Object 3 should be placed closer to object 1.

 

Object 3 should be placed on a halfway between object 2 and object 1.

 

Object 3 should be placed closer to object 2.

 

Solution

I think that Object 3 should be placed closer to object 2.

6 0
3 years ago
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