I believe the correct response would be B. It would decrease.
Use this formula to find your answer...
Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms
frequency (f)=1/( Time period).
Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.
Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.
To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
The drift velocity is given by the equation:
Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,
The hall voltage is given by
Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,
Answer:
4.535 N.m
Explanation:
To solve this question, we're going to use the formula for moment of inertia
I = mL²/12
Where
I = moment of inertia
m = mass of the ladder, 7.98 kg
L = length of the ladder, 4.15 m
On solving we have
I = 7.98 * (4.15)² / 12
I = (7.98 * 17.2225) / 12
I = 137.44 / 12
I = 11.45 kg·m²
That is the moment of inertia about the center.
Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that
τ = 11.453 kg·m² * 0.395 rad/s²
τ = 4.535 N·m
