A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is
Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:
and substituting t = 75 seconds, we find
In degrees, it is
So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by
where we have
r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find
Answer: Move the small car so it appears on the left side of the lens.
Explanation:
Because the lens is reflective the small car would apear on the same side as the normal car.
Hope this helps :)
Frequency represents the number of complete oscillations in one second. it is measured in Hertz (Hz). Electromagnetic waves are waves which do not require a material media for transmission. They travel with a speed of light.
The speed (m/s) of a wave is given by frequency (Hz) × Wavelength (m)
Speed is 300,000 km/sec or 300,000,000 m/s and the wavelength is 300,000 km or 300,000,000 m.
Frequency = speed÷ wavelength
= 300000000 ÷ 300000000 = 1
Therefore, the frequency of the wave is 1Hz
Answer:
F' = (3/2)F
Explanation:
the formula for the electric field strength is given as follows:
E = F/q
where,
E = Electric Field Strength
F = Force due to the electric field
q = magnitude of charge experiencing the force
Therefore,
F = E q ---------------- equation (1)
Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:
F' = (E/2)(3 q)
F' = (3/2)(E q)
using equation (1)
<u>F' = (3/2)F</u>
Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;
Therefore, the frequency of this mode of vibration is 138.87 Hz
