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Andrew [12]
3 years ago
8

. Rohit had only one light source in his room. He was told by his friend that he can make the room more lighted by using mirror.

What type of mirror can be used for the purpose?
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0

Answer:

very smooth and polished glass mirror

Explanation:

The best type of mirror for this would be a very smooth and polished glass mirror. A metal mirror will also work but it would need to be extremely polished. This would allow light to efficiently be reflected across the room, thus effectively adding an "extra" light source to the room. If the mirror is not smooth then the light may become distorted and not reflect properly as it is supposed to. This may defeat the entire purpose of the mirror for this scenario.

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Consider an electric dipole in a uniform electric field. In which orientation does the dipole-field system have the greatest pot
Gekata [30.6K]

Answer: when the dipole moment and electric field are parallel

Explanation: The formulae that relates the potential energy of a dipole, dipole moment and strength of electric field is given as

u = p * E cosθ

Where, u= dipole potential energy, p = dipole moment, E = strength of electric field.

The expression is at maximum when θ = 0 (cos 0° = 1)

Hence the function for potential energy will be greatest when θ = 0° which implies that the dipole moment and strength of electric field are parallel to each other.

5 0
3 years ago
How would you find the horizontal net force for the free body diagram below
tiny-mole [99]

Answer:

Add Ff from Fa

Explanation:

Fnet = sum of all force

horizontal net force = Ff + Fa

7 0
2 years ago
A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
jeka94

Answer:

a)  t_l - t_r = 12.54 us

b)  (t_l - t_r) / T = 0.0157  

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

                                      d_l = sqrt ( 2.6^2 + 0.15^2)

                                      d_l = sqrt ( 6.7825 )

                                      d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

                                      t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

                                      t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)  

                                      t_l - t_r = ( 0.0043 / 343 )

                                      t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

                                      T = 1 / f

                                      T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

                                      (t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

                                      (t_l - t_r) / T = 0.0157  

3 0
3 years ago
The surface tension of isopropanol in air has a value of 23.00 units and the
Y_Kistochka [10]

Answer:

It's A & C

Explanation:

:p

7 0
3 years ago
Read 2 more answers
Two Polaroids are aligned so that the initially unpolarized light passing through them is a maximum. At what angle should one of
steposvetlana [31]

To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

I = I_0 cos^2\theta

Where,

I_0= Initial Intensity

I = Final Intensity after pass through the polarizer

\theta= Angle between the polarizer and the light

Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as

\frac{I}{I_0} = \frac{1}{2}

Using the Malus Law we have,

I = I_0 cos^2\theta

cos^2\theta = \frac{I}{I_0}

cos^2\theta = \frac{1}{2}

\theta = cos^{-1}(\frac{1}{2})^2

\theta = 75.52\°

Angle with respect to maximum is 90-75.52 = 14.48\°

8 0
2 years ago
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