Answer:
2 × 10⁶
Explanation:
Data provided in the question:
Cavity length, L = 
Oscillation frequency, 
= 9.0 × 10¹⁴ Hz
Now,
we know,
here,
c is the speed of light = 3 × 10⁸ m/s
= Wavelength of mode m inside the laser cavity
m is the cavity mode number
Thus,
or
= 
× 10⁻⁶
Also,
Therefore,
m × × 10⁻⁶ = 2 ×
or
m = 2 × 10⁶
Answer:
Check Explanation
Explanation:
Let the speed of the drone relative to earth be u = ?
Let the speed of the drone relative to the starship be u' = 0.5c
Let the speed of the starship relative to earth be v = 0.8c
In the theory of relativity,
The 3 speeds are related thus
u = (u' + v)/[1 + (u'v/c²)]
u = (0.5c + 0.8c)[1 + ((0.8c × 0.5c)/c²)]
u = 1.3c/[1 + (0.4c²/c²)]
u = 1.3c/1.4 = 0.9286 c = 0.93 c = 93% c
Answer:
No, the apple will reach 4.20041 m below the tree house.
Explanation:
t = Time taken
u = Initial velocity = 2.8 m/s
v = Final velocity = 0
s = Displacement
g = Acceleration due to gravity = -9.81 m/s² = a (negative as it is going up)
Equation of motion
The height to which the apple above the point of release will reach is 0.39959 m
From the ground the distance will be 1.3+0.39959 = 1.69959 m
Distance from the tree house = 5.9-1.69959 = 4.20041 m
No, the apple will reach 4.20041 m below the tree house.
The values in the option do not reflect the answer.
Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C
Answer:
0.83 ω
Explanation:
mass of flywheel, m = M
initial angular velocity of the flywheel, ω = ωo
mass of another flywheel, m' = M/5
radius of both the flywheels = R
let the final angular velocity of the system is ω'
Moment of inertia of the first flywheel , I = 0.5 MR²
Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²
use the conservation of angular momentum as no external torque is applied on the system.
I x ω = ( I + I') x ω'
0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'
0.5 x MR² x ωo = 0.6 MR² x ω'
ω' = 0.83 ω
Thus, the final angular velocity of the system of flywheels is 0.83 ω.
