Answer:
time taken by the wave to reach the person is 0.2 s
Explanation:
As we know that the speed of the wave is given as

here we know that the wavelength of the wave is


now speed of the wave is given as


Now time taken by the wave to reach 5 m distance is



Answer:
16613 m/s
Explanation:
Given that
mass of the fly, m = 0.55 g = 0.55*10^-3 kg
Kinetic Energy of the fly, E = 7.6*10^4 J
Speed of the fly, v = ? m/s
We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.
The Kinetic Energy, KE of any object is represented by the formula
KE = 1/2 * m * v²
If we substitute the values in the relation, we have,
7.6*10^4 = 1/2 * 0.55*10^-3 * v²
v² = (15.2*10^4) / 0.55*10^-3
v² = 2.76*10^8
v = √2.76*10^8
v = 16613 m/s
Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J
see i was trying to figure out the answer but i didn't understand it so i took the time to research and work it out but i still didn't understand i found one that was close to it and i got the same one as the other person which is D but idk if it is that type of question if it is than it is d if not then idk
Answer:
The maximum value of the induced magnetic field is
.
Explanation:
Given that,
Radius of plate = 30 mm
Separation = 5.0 mm
Frequency = 60 Hz
Suppose the maximum potential difference is 100 V and r= 130 mm.
We need to calculate the angular frequency
Using formula of angular frequency

Put the value into the formula


When r>R, the magnetic field is inversely proportional to the r.
We need to calculate the maximum value of the induced magnetic field that occurs at r = R
Using formula of magnetic filed

Where, R = radius of plate
d = plate separation
V = voltage
Put the value into the formula


Hence, The maximum value of the induced magnetic field is
.
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.