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2 years ago

Suppose that a barometer was made using oil with rho=900 kg/m3. What is the height of the barometer at atmospheric pressure?

Physics

1 answer:

rewona [7]2 years ago

5 0

Hey there!

The pressure under a liquid column can be , calculated using the following formula :

P = p x g x h

P atm = 1.013 x 10⁵ Pa

g = 9.8 m/s²

h = ?

h = P / ( p x g ) =

h= ( 1.013 x 10⁵ Pa ) / ( 900 x 9.8 ) =

h = ( 1.013 x 10⁵ ) / ( 8820 ) =

h = 11.48 m ≈ 11.50 m

Hope this helps!

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Which formula is used to find fluctuation of the shape of body

Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0

2 years ago

6. Starting from 1.5 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate o

loris [4]

For the answer to the question above,

<span>To be 0.1 miles away from the check point ,
the car has to travel 1.4 miles OR 1.6 miles. </span>

53 miles = 60 minutes

1.4 miles = 1.4 / 53 X 60 = 1.5849056 minutes OR 95.1 seconds

<span>1.6 miles = 1.6 /53 X 60 = 1.8113207 minutes OR 108.7 seconds
</span>So the answer is <span>95.1s and 108.7s
I hope my answer helped you</span>

7 0

2 years ago

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direc

Alborosie

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is $6.21 \times 10^{-3}$ N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

Q = $6.50 \times 10^{-6} C$

E = 1300 N/C in the +x direction

B = 1.02 T in the +z direction

and, $F_{net} = 6.25 \times 10^{-3} N$ in the +x direction

Also, $F_{net} = F_{E} - F_{b}$

= qE - qvB

Now, we will calculate the value of v as follows.

v = $(\frac{1}{B}) \times (E - \frac{F_{net}}{q})$

= $(\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})$

v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for $F_{net}$ .

Using the right hand rule where:

your right thumb goes toward the $F_{net}$ , then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that $v_{x}, v_{y}, v_{z}$ = 0, -(458.507), 0.

8 0

2 years ago

Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per

vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work and amount of entropy and magnitude

solution

first we calculate work i.e heat transfer - work = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C .................1

so first we get some value from steam table with the help of 100 bar @360°C and 1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change = specific entropy ( 100 bar @360°C) - specific entropy ( 1 bar @160°C )

put these value we get

entropy change = 7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum theoretical work = specific internal energy - 2038.3

maximum theoretical work = 2728 - 2038.3

maximum theoretical work is 689.4 kJ/kg

3 0

2 years ago

How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)

I am Lyosha [343]

<h2>Option C is the correct answer.</h2>

Explanation:

We need to find how many calories is 1 BTU.

Given

1 BTU = 1054 J

1 calorie = 4.186 J

So we have

1 BTU = 4.186 x 251.79 J

1 BTU =251.79 calorie

1 BTU = 252 calorie.

Option C is the correct answer.

3 0

2 years ago