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3 years ago

6

Why do you see objects when you shine a flashlight in a dark room?

1 answer:

anzhelika [568]3 years ago

8 0

Explanation:

We see objects in a dark room due to the emission of light photons which are sensitive to our eyes.

Darkness is simply a terminology used to describe the absence of light. Visible light to human is a component of the electromagnetic spectrum. Our eyes have receptors that picks the photons which light releases.

The particles of light are called photons.

When a torch is shines. It produces light due to emission of light by excitation of the metal in the bulb.
The light is in form of photons that spreads and travels in all direction in a room.
When they impinge on other bodies, they cause them to excite and release their own photons
Photons from the different parts helps to depict an object as it is.
The photo-receptors in our eyes picks up the photon and interprets it in our visual faculty.

learn more:

light waves brainly.com/question/8032392

#learnwithBrainly

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You are 12 miles north of your base camp when you begin walking north at a speed of 2mph. what is your location, relative to you

Leona [35]

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3 years ago

A thin film of oil of thickness t is floating on water. The oil has index of refraction no = 1.4. There is air above the oil. Wh

kkurt [141]

Answer:

t = 120.5 nm

Explanation:

given,

refractive index of the oil = 1.4

wavelength of the red light = 675 nm

minimum thickness of film = ?

formula used for the constructive interference

$2 n t = (m+\dfrac{1}{2})\lambda$

where n is the refractive index of oil

t is thickness of film

for minimum thickness

m = 0

$2 \times 1.4 \times t = (0+\dfrac{1}{2})\times 675$

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3 years ago

Give three examples from your life of magnetic force.

Margaret [11]

Headphones, refrigerator magnets, and compasses

Hope that was helpful.

4 0

3 years ago

Read 2 more answers

A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy

Vsevolod [243]

Answer:

$v = 4.2 \ m/s$

Explanation:

Given data:

Mass of the paper clip, $m = 1.5 \ g = 0.0015 \ kg$

Kinetic energy, $K = 0.013 \ \rm J$

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

$K = \frac{1}{2}mv^{2}$

$0.013 = 0.5 \times 0.0015 \times v^{2}$

$\Rightarrow \ v = 4.16 \ m/s$

$v = 4.2 \ m/s.$ (rounding to nearest tenth)

3 0

3 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago