As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "
Doppler's effect."
Now the general formula of the Doppler's effect is:
-- (A)
Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.
Where,
g = Speed of sound = 340m/s.
= Velocity of the receiver/observer relative to the medium = ?.
= Velocity of the source with respect to medium = 0 m/s.
= Frequency emitted from source = 400 Hz.
= Observed frequency = 408Hz.
Plug-in the above values in the equation (A), you would get:
Solving above would give you,
= 6.8 m/s
The correct answer = 6.8m/s
<h2>For Second Solid Lumped System is Applicabe</h2>
Explanation:
Considering heat transfer between two identical hot solid bodies and their environments -
- If the first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air than for second solid, the lumped system analysis more likely to be applicable
- The reason is that a lumped system analysis is more likely to be applicable in the air than in water as the convection heat transfer coefficient so that the Biot number is less than or equal to 0.1 that is much smaller in air
Biot number = the ratio of conduction resistance within the body to convection resistance at the surface of the body
∴ For a lumped system analysis Biot number should be less than 0.1
Answer:
The final velocity of the bullet is 9 m/s.
Explanation:
We have,
Mass of a bullet is, m = 0.05 kg
Mass of wooden block is, M = 5 kg
Initial speed of bullet, v = 909 m/s
The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,
So, the final velocity of the bullet is 9 m/s.
Answer:
Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m
Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m
Explanation:
Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.
H = (v₀² Sin²θ)/2g
v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s
θ = 75°, g = 9.8 m/s²
H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m
Range of projectile
R = v₀² (sin2θ)/g
R = 250² (sin2×75)/9.8
R = 250² (sin 150)/9.8 = 3188.8 m
Height of mountain = 1.80 × 10³ = 1800 m
Maximum height of projectile = 2975.2 m
Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m
Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m
Range of projectile = 3188.8 m
Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m
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.
