Answer:
a) For P: 
For Q: 
b) For P:

for Q:

c) As the distance from the axis increases then speed increases too.
Explanation:
a) Assuming constant angular acceleration we can find the angular speed of the wheel dividing the angular displacement θ between time of rotation:

One rotation is 360 degrees or 2π radians, so θ=2π

Angular acceleration is at every point on the wheel, but speed (tangential speed) is different and depends on the position (R) respect the rotation axis, the equation that relates angular speed and speed is:

for P:

for Q:

b) Centripetal acceleration is:

for P:

for Q:

c) As seen on a) speed and distance from axis is
because ω is constant the if R increases then v increases too.
Answer:
10 GHz
Explanation:
Applying,
v = λf.................... Equation 1
Where v = speed of microwave, λ = wavelength, f = frequency.
make f the subject of the equation
f = v/λ................ Equation 2
Note: Microwave is an electromagnetic wave, and all electromagnetic wave have the same speed, which is 3×10⁸ m/s
From the question,
Given: λ = 3 cm = 0.03 m
Constant; v = 3×10⁸ m/s
Substitute these values into equation 2
f = 3×10⁸/(0.03)
f = 10¹⁰ Hz
f = (10¹⁰/10⁹) GHz
f = 10 GHz
Sound level at distance of 15 m is given as 20 dB
so intensity at this distance is given as



now if we move closer to some some distance the sound level is now 50 dB
now the intensity is given as



now we know that



so now the distance from friend must be 47 cm
Answer:
88.8 m/s= Speed of wave propagation in the required mode.(3 loops)
Explanation:
When there are 3 loops.
the total length = L = 3 λ /2
⇒ λ = 2 L / 3 = 2 ( 1.11 ) / 3 = 0.74 m
Velocity = v = f λ = (120)(0.74) = 88.8 m/s
The properties of the wave don't determine its speed. The properties of the medium do. You can FIND the speed by measuring the wave's frequency and wavelength.