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

Answers

Answer 1

Answer:

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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The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7 T · N/A and the speed of light is 2.99792 × 108 m/s. Find the magnitude of Em for the electromagnetic waves at the top of the atmosphere. Answer in units of N/C.

Answers

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed ,E=Electric filed

Power P = I A

A=Area=4πr² ,I=Intensity

$I=(CB^2)/(2\mu_0)$

$I=(CE^2)/(2\mu_0 C^2)$

$E=\sqrt{{2I\mu_0 C}}$

$E=\sqrt{{2* 1150* 4\pi * 10^(-7)(2.99792* 10^8)}}$

E=930.84 N/C

Therefore answer is 930.84 N/C

Final answer:

To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.

Explanation:

To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.

Subbing in the given values, we can compute Em as:

Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]

This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.

Learn more about Electromagnetic Waves here:

brainly.com/question/29774932

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Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

Answers

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

Assume the electric field has a magnitude of 59.0 m V , and the particle has a net positive charge of Q=+5.40 C. Calculate the magnitude of the electric potential difference, \vert \Delta V \vert = \vert V_B - V_A \vert∣ΔV∣=∣V B −V A ∣.

Answers

Answer:

V = 1.69 * 10^6 V

Explanation:

Parameters given:

Electric field, E = 59V/m

Charge, q = 5.40C

We need to first find the distance between the electric charge and the point of consideration to be able to find the Electric potential difference.

Electric field is given as:

E = (kq/r^2)

k = Coulombs constant

=> r^2 = kq/E

=> r^2 = (9 * 10^9 * 5.4) / 59

r^2 = 8.2 * 10^8

r = 2.84 * 10^4 m

We can now find the Electric Potential by using:

V = kq/r

Hence,

V = (9 * 10^9 * 5.4) / (2.84 * 10^4)

V = 1.69 * 10^6 V

Draw a ray diagram for an object placed more than two focal lengths in front of a converging lens.

Answers

Answer:

Explanation:

There is a convex lens M N is placed. An object AB is placed at a distance more than two focal lengths of the lens.

A ray of light is starting from point A and parallel to the principal axis, then after refraction it goes from the focus.

Another ray which goes through the optical centre of the lens becomes undeviated after refraction.

The two refracted rays meet at the point A', So A'B is the image of AB.

The nature of image is real, inverted and diminished.

A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.