Answer:
<u>Amplitude - remains the same</u>
<u>Frequency - increases</u>
<u>Period - decreases</u>
<u>Velocity - remains the same.</u>
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Explanation:
The amplitude of the wave remains the same since you are not changing the distance your hand moves and the amplitude of the wave depends on how much distance your hand covers while moving.
The frequency of your wave increases since now you are moving your hand more number of times in the same period i.e. your hand is moving faster in one second. So, the frequency of your wave increases.
The period is the time taken by the wave to travel a certain distance. Since your hand is now moving faster, the wave will travel faster and will take less time to cover the same distance hence, we can say that its period will decrease.
The velocity of a wave depends on the medium in which it is travelling. Your wave was previously travelling in air and the new wave is also travelling in the same medium so the velocity of the wave remains unchanged.
Answer:
the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.
Explanation:
Given;
diameter of the ceiling fan, d = 90 cm = 0.9 m
angular speed of the fan, ω = 64 rpm
time taken for the fan to stop, t = 28 s
The distance traveled by the ceiling fan when it comes to a stop is calculated as;

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.
Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R = 
where
R = resistance

f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:

where
x = separation between the parallel plates
Thus
C ∝ 
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,

Also,
Z ∝ I
Therefore,




Solving the above eqn:

Answer:
The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Explanation:
Thickness of the wall is L= 20cm = 0.2m
Thermal conductivity of the wall is K = 2.79 W/m·K
Temperature at the left side surface is T₁ = 50°C
Temperature of the air is T = 22°C
Convection heat transfer coefficient is h = 15 W/m2·K
Heat conduction process through wall is equal to the heat convection process so

Expression for the heat conduction process is

Expression for the heat convection process is

Substitute the expressions of conduction and convection in equation above


Substitute the values in above equation

Now heat flux through the wall can be calculated as

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²