Answer:
here, v = 0, a = - 9.8 m/sec^2, t = 3 sec
v = u + at
0 = u + (-9.8) * 3
0 = u - 29.4
u = 29.4 m/sec
here, u = 29.4m/s, t = 3sec, a= -9.8m/s^2
s = ut + 1/2 a t^2
s = 29.4 * 3 + 1/2 * - 9.8 * 3^2
s = 88.2 - 4.9 * 9
s = 88.2 - 44.1
s = 44.1 m
here, u = 0, a = 9.8m/s^2, t = 1 sec, s = ?
position after 4sec :
s = ut+ 1/2 at^2
s = 0 * 1 + 1/2 * 9.8 * 1^2
s = 4.9m
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Monochromatic light strikes a metal surface and electrons are ejected from the metal. if the intensity of the light is increased, the ejection rate will increase, but t<span>he maximum energy remains the same</span> .
Velocity and speed of a moving body become identical when it tends to move in a straight line.
For the current to reach half its final value:
Answer
given,
Electric field,E = 490 N/C
time, t = 54 ns
for electron
Mass of electron me = 9.1 x 10⁻³¹ kg
Charge of electron e = -1.6 x 10⁻¹⁹ C
electrostatic force
F = E q
F = 490 x 1.6 x 10⁻¹⁹
F = 784 x 10⁻¹⁹ N
now, using newton second law
a = 8.62 x 10¹² m/s²
using equation of motion
v = u + a t
v = 0 + 8.62 x 10¹² x 54 x 10⁻⁹
v = 4.65 x 10⁵ m/s
velocity of electron is equal to v = 4.65 x 10⁵ m/s
For Proton
Mass mp = 1.67 x 10⁻²⁷ kg
Charge p = 1.6 x 10⁻¹⁹ C
Electric field E = 490 V/C
from above solution
F = 784 x 10⁻¹⁹ N
now, acceleration
a = 4.69 x 10¹⁰ m/s²
using equation of motion
v = u + a t
v = 0 + 4.69 x 10¹⁰ x 54 x 10⁻⁹
v = 4.65 x 10³ m/s
velocity of electron is equal to v = 4.65 x 10³ m/s