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Serhud [2]
3 years ago
7

A camera flash, which throws a parallel beam of light, has a mirror behind the light bulb. What kind of mirror is it likely to b

e?
Physics
2 answers:
weqwewe [10]3 years ago
7 0

Answer:

It is concave mirror.

Explanation:

If a light source be placed at the focus of concave mirror, the rays of light falling on the concave mirror and then after reflection, the rays of light are parallel to each other.

WINSTONCH [101]3 years ago
3 0
For Plato, the correct answer is Concave mirror :)
You might be interested in
Two equal masses are attached to separate identical springs next to one another. One mass is pulled so its spring stretches 20 c
wlad13 [49]

Answer:

Both reaches equilibrium simultaneously.

Explanation:

Given that,

First stretch spring= 20 cm

Second stretch spring = 10 cm

Two equal masses are suspended from identical spring.

Their forces constants are equal time period for oscillation

Using formula of time period

T=2\pi\sqrt{\dfrac{m}{k}}

Where, T = time period

m = mass

k = spring constant

Time period independent of amplitude.

Hence, Both reaches equilibrium simultaneously.

7 0
3 years ago
Read 2 more answers
A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from
Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of  incidence  beam striking the water is \theta = 49.63^o

c  the angle of  refraction  beam striking the water is r = 59.7^o

d The angle the refracted beam make with respect to the horizontal is = 30.3^o

e The height of the target above sea level is  

                   h= 125.05m

Explanation:

From the diagram we see that the angle of the beam striking the water is

                   tan \theta = \frac{100}{85}

                        \theta = tan^{-1}(\frac{100}{85} )

                           = 49.63^o

According to Snell's law

                   \mu_{water} *sin(i) = \mu_{air}  *sin(r)

Where \mu_{water } is the refractive index of water =  1.333

           i is the angle of incidence

          \mu_{air} is the refractive index of air  = 1

            r is the angle of refraction

 Substituting values accordingly

          1.33 * sin (40.37) = 1 * sin(r)

    Making r the subject of the formula

                       r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})

                          = 59.7^o

looking at the diagram we can see that to  obtain the angle the refraction beam makes with the horizontal   by subtracting the angle refraction from 90°

                 i.e  90 -59.7 = 30.3 °

From the diagram we see that the height  target above sea level can be obtained by this relation

                   tan \theta = \frac{h}{214}\\

Where h is the is the height

                   tan(30.3) = \frac{h}{214}

                         h = 214 * tan (30.3)

                            =125.05m

           

4 0
3 years ago
(DUE IN FIVE MINUTES, QUICK)
goldfiish [28.3K]

Answer:

Well you would be flouting out in space so it will change your weight a lot because you can't stop moving.

8 0
3 years ago
Read 2 more answers
Mr Ndlovu works in centurion and works at Rosebank.It takes him 13minutes to get to work and he needs to be there at 08h15
VARVARA [1.3K]

Answer:

8:02 no less

Explanation:

8:15 - 13mins = 8:02

6 0
3 years ago
Question 5 At 12:00 pm, a spaceship is at position ⎡⎣324⎤⎦ km ⎣ ⎢ ⎡ ​ 3 2 4 ​ ⎦ ⎥ ⎤ ​ km away from the origin with respect to so
Anettt [7]

Answer:

[1, 6, -2]

Explanation:

Given the following :

Initial Position of spaceship : [3 2 4] km

Velocity of spaceship : [-1 2 - 3] km/hr

Location of ship after two hours have passed :

Distance moved by spaceship :

Velocity × time

[-1 2 -3] × 2 = [-2 4 -6]

Location of ship after two hours :

Initial position + distance moved

[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]

= [3-2, 2+4, 4-6] = [1, 6, -2]

4 0
3 years ago
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