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
Answer:
Explanation:
The expression for the calculation of the enthalpy change of a process is shown below as:-
Where,
is the enthalpy change
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass of water = 2.4 kg
Specific heat = 4.18 J/g°C
So,
Heat Supplied 
where 



Answer:
Explanation:
a )
one kg of coal gives energy of 27 x 10⁶ J
75 kg of coal gives energy of 27 x 10⁶ x 75 J
So rate which energy is coming out of coal per second
= 27 x 10⁶ x 75 J
= 2025 x 10⁶ J /s
2025 million watts .
b ) energy output = 800 million watts
efficiency = (800 / 2025) x 100
= 39.5 % .
Answer:
230
Explanation:
= Rotational speed = 3600 rad/s
I = Moment of inertia = 6 kgm²
m = Mass of flywheel = 1500 kg
v = Velocity = 15 m/s
The kinetic energy of flywheel is given by

Energy used in one acceleration

Number of accelerations would be given by

So the number of complete accelerations is 230
Answer:
E/4
Explanation:
The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
Where;
E is the electric field
σ is the surface charge density
ε₀ is the electric constant.
Formula to calculate σ is;
σ = Q/A
Where;
Q is the total charge of the sheet
A is the sheet's area.
We are told the elastic sheet is a square with a side length as d, thus ;
A = d²
So;
σ = Q/d²
Putting Q/d² for σ in the electric field equation to obtain;
E = Q/(2ε₀d²)
Now, we can see that E is inversely proportional to the square of d i.e.
E ∝ 1/d²
The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.
From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;
E_new = E/4