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love history [14]
2 years ago
14

Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo

ats partly in the oil and partly in the water. The oil completely covers the block. How far below the interface between the two liquids is the bottom of the block? Answer in units of m.
Physics
1 answer:
Blizzard [7]2 years ago
3 0

Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

wood

gh−ρ

oil

g(h−x)−ρ

water

gx=0

\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ

wood

h−ρ

oil

(h−x)−ρ

water

x=0

(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0

x=2.54\ cmx=2.54 cm

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sertanlavr [38]

Answer:

c

Explanation:

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5 0
2 years ago
Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t
Nataliya [291]

Answer:

 Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

                                     I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

                                     Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

                                     Q = cos^-1 ( sqrt (0.464))

                                     Q = cos^-1 (0.6811754546)

                                     Q = 47.06 degrees

                                   

8 0
3 years ago
Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected
Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance C= 20\times10^{-2}\ F

Inductance L=20\times10^{-3}\ H

Resistance R=  50\ Omega

We need to calculate the impedance

Using formula of impedance

z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}

z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}

z=50.00

We need to calculate the phase angle

Using formula of phase angle

\theta=\cos^{-1}(\dfrac{R}{z})

\theta=\cos^{-1}(\dfrac{50}{50.00})

\theta=0.0180\ rad

Hence, The phase angle is 0.0180 rad.

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3 years ago
Which of these will always produce a magnetic field?
vovikov84 [41]

Answer:

Technically everything has somewhat of a magnetic field. I guess

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3 years ago
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