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Digiron [165]
2 years ago
7

You position two plane mirriors at right angles to each other a light ray strikes one mirrior at an angle of 60 to the normal an

d reflects towards the second mirrior what is the angle of reflection off the second mirrior
Physics
1 answer:
Dafna1 [17]2 years ago
4 0

Answer:

30 degrees

Explanation:

Reflects off of mirror 1  at   60 degrees....this makes it incident to second mirror at 30 degrees ....then angle of reflection equals this angle of incidence = 30 degrees

See atached diagram

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Arjun told to his friend that " Q can form a virtual image larger than the object by refraction." What is "Q"?
nadezda [96]

Q is a concave mirror.

Explanation:

The image formed by a concave mirror is observed to be virtual, erect and larger than the object.

7 0
2 years ago
Particle 1 has mass 4.6 kg and is on the x-axis at x = 5.7 m. Particle 2 has mass 7.2 kg and is on the y-axis at y = 4.2 m. Part
Iteru [2.4K]

To solve this problem it is necessary to apply the concepts related to the Gravitational Force, for this purpose it is understood that the gravitational force is described as

F_g = \frac{Gm_1m_2}{r^2}

Where,

G = Gravitational Universal Force

m_i = Mass of each object

To solve this problem it is necessary to divide the gravitational force (x, y) into the required components and then use the tangent to find the angle generated between both components.

Our values are given as,

m_1 =4.6 kg\\m_2 = 7.2 kg\\m_3 = 2.6 kg\\r_1 = 5.7 m\\r_2 = 4.2 m

Applying the previous equation at X-Axis,

F_x = \frac{Gm_1m_3}{R_{1}^2}\\F_x = \frac{6.67*10^{-11}*4.6*2.6}{5.7^2}\\F_x = 2.46*10^{-11}N

Applying the previous equation at Y-Axis,

F_y = \frac{Gm_2m_3}{R_2^2}\\F_y = \frac{6.67*10^{-11}*7.2*2.6}{4.2^2}\\F_y = 7.08*10^{-11} N

Therefore the angle can be calculated as,

tan\theta = \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{7.08*10^{-11}}{2.46*10^{-11}}\\\theta = 71\°

Then in the measure contrary to the hands of the clock the Force in the particle 3 is in between the positive direction of the X and the negative direction of the Y at 71 ° from the positive x-axis.

5 0
4 years ago
PLEASE HELP (100)
morpeh [17]

Answer:

The work done will be W=55.27\: J

Explanation:

The work equation is given by:

W=F\cdot x

Where:

F is the force due to gravity (weight = mg)

x is the length of the ramp (3 m)

Now, the force acting here is the component of weight in the ramp direction, so it will be:

F_{x-direction}=mgsin(28)

Therefore, the work done will be:

W=mgsin(28)*3

W=4*9.81*sin(28)*3

W=55.27\: J

I hope it helps you!

3 0
3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
PLEASE ANSWER ASAP
vivado [14]
Reduce the rate of global warming
4 0
4 years ago
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