Answer:
VB − VA = g tAB & (VA + VB)/2 = h / tAB
Explanation:
s = h = Displacement
tAB = t = Time taken
VA = u = Initial velocity
VB = v = Final velocity
a = g = Acceleration due to gravity = 9.8 m/s²
Hence, the equations VB − VA = g tAB & (VA + VB)/2 = h / tAB will be used
Half the potential difference of the the1-µF
A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.
The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.
Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.
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fredd [130]
Answer:
the distance covered by the body in 5th second =u+a(n- 1/2)=7+4(5- 1/2)
=7+4×9/2
=7+18
=25m
Answer:
Explanation:
Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km
Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km
Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite
Orbital potential energy of a satellite A = - GMm / Ra
Orbital potential energy of a satellite B = - GMm / Rb
PE of satellite B /PE of satellite A
= Ra / Rb
= 12740 / 25480
= 1 / 2
b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same
KE of satellite B /KE of satellite A
= 1 / 2
c ) Total energy will be as follows
Total energy = - PE + KE
- P E + PE/2
= - PE /2
Total energy of satellite B / Total energy of A
= 1 / 2
Satellite B will have greater total energy because its negative value is less.
Answer:
The velocity of the light will be 1.0c only
Explanation:
The velocity of the light measured in the case given in question will be 1.0c only.
This is due to the fact that the velocity of light is never relative. The velocity of the light is maximum
The velocity of the light cannot be scaled down in no case
Thus, the velocity of the light remains as constant.
Hence, the velocity of the light measured will be 1.0c although the ships have relative velocity.