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Free_Kalibri [48]
3 years ago
13

Derive the dimension of coefficient of linear expansivity

Physics
2 answers:
Monica [59]3 years ago
5 0
Q1. A metal rod is of length 64.576 cm at a temperature 90°C whereas the same metal rod has a length of 64.522 cm at a temperature 12°C. Calculate the coefficient of linear expansion.
Reil [10]3 years ago
3 0

Answer:

The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.

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A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic
xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2

 =2.5m

on the axis of x at x = 2.5

r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}

r = 2.5m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T

(Along z direction)

B)r = 5.00m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2

 =5.00m

on the axis of x at x = 5.0

r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}

r = 5.00m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T

(Along x direction)

7 0
3 years ago
How is the combined gas law used to calculate changes in pressure, temperatures, and/or volume for a fixed amount of gas?
nalin [4]

Answer:

Let's start by considering the ideal gas law:

pV=nRT

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature

This equation can also be rewritten as

\frac{pV}{T}=nR

Now, if we consider a fixed amount of gas, this means that the number of moles (n) is constant. So we can rewrite the equation as

\frac{pV}{T}=const.

And therefore, if we consider a gas undergoing a certain transformation from 1 to 2, we can write

\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}

where 1 indicates the conditions of the gas at the beginning and 2 the conditions of the gas after the process. So, the change in pressure/temperature/volume of the gas can be found by using this equation.

5 0
3 years ago
The velocity of a car traveling east increases from 4 m/s to 10 m/s in 20 second what is acceleration of the car?
Grace [21]
A=4-10 ÷ 20
a=6÷20
a=0.3
5 0
3 years ago
A student is eating french fries and needs ketchup. He shakes the ketchup bottle as it is turned upside down but the ketchup won
Harrizon [31]

Answer:

I'm thinking Newton's first law of motion.

Explanation:

"An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force."

8 0
3 years ago
If a ball has a mass of 5 kg and 100 J of KE, what is its velocity?
Yanka [14]

{\fcolorbox{white}{lightgreen}{\bf{\textcircled{$\checkmark$}}{Verified\:answer}}}

  • Mass of ball=m=5kg
  • Kinetic energy=KE=100J
  • Velocity=v=?

We know

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

\\ \sf\longmapsto 100J=\dfrac{1}{2}5\times v^2

\\ \sf\longmapsto v^2=100\times \dfrac{2}{5}

\\ \sf\longmapsto v^2=20(2)

\\ \sf\longmapsto v^2=40

\\ \sf\longmapsto v=\sqrt{40}

\\ \sf\longmapsto v=6.2m/s

8 0
2 years ago
Read 2 more answers
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