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2 years ago

If the light strikes the first mirror at an angle θ1, what is the reflected angle θ2? express your answer in terms of θ1.

Physics

1 answer:

Alexus [3.1K]2 years ago

3 0

Answer:

θ₂ = 90° - θ₁

Explanation:

When the light falls on a mirror it bounces back. This is know as reflection. The incident angle is equal to the angle of reflection.

Here, the light strikes the mirror at an angle = θ₁

To find the angle of reflection we first need to understand angle of incidence. The angle of incidence is the angle made between the incident ray and normal. Normal is an imaginary line drawn perpendicular line on the boundary of the mirror.

Since the light strikes the mirror at angle of θ₁, which is the angle between light ray and the mirror.

Angle of incidence = 90° - θ₁.

Thus, angle of reflection, θ₂ = 90° - θ₁

You might be interested in

What is the magnitude of the relative angle φ

melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0

2 years ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0

2 years ago

A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer

Mekhanik [1.2K]

The initial volume of the gas is
$V_i = 600 cm^3$
while its final volume is
$V_f = 200 cm^3$
so its variation of volume is
$\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3$

The pressure is constant, and it is
$p=400 kPa = 400 \cdot 10^3 Pa$

Therefore the work done by the gas is
$W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J$
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
$Q=+100 J$

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
$\Delta U = Q-W = 100 J - (-160 J)=+260 J$
where the positive sign means the internal energy of the gas has increased.

7 0

2 years ago

g A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0

Margarita [4]

Answer:

Explanation:

If a particle move with time and expressed according to the formula:

f(t) = 0.01t⁴ − 0.03t³

a) Velocity is the change in motion of the particle with respect to time and it is expressed as;

$v(t) =\frac{d(f(t))}{dt}$

$v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2$

Hence the velocity of the particle at time t is $v(t) = 0.04t^3 - 0.09t^2$

b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:

$v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05$

Hence the velocity after 1second is -0.05

c) The particle is at rest when when the time is zero.

Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second

6 0

2 years ago

A person walks 25 m west and then 45 m at the angle of 60 degrees north of east what is the magnitude of the total displacement?

expeople1 [14]

To solve this question, we need to use the component method and split our displacements into their x and y vectors. We will assign north and east as the positive directions.

The first movement of 25m west is already split. x = -25m, y = 0m.

The second movement of 45m [E60N] needs to be split using trig.
x = 45cos60 = 22.5m
y = 45sin60 = 39.0m

Then, we add the two x and two y displacements to get the total displacement in each direction.

x = -25m + 22.5m = -2.5m
y = 0m + 39.0m

We can use Pythagorean theorem to find the total displacement.
d² = x² + y²
d = √(-2.5² + 39²)
d = 39.08m

And then we can use tan to find the angle.
inversetan(y/x) = angle
inversetan(39/2.5) = 86.3

Therefore, the total displacement is 39.08m [W86.3N]

8 0

2 years ago