Answer:
Explanation:
q = 2e = 3.2 x 10^-19 C
mass, m = 6.68 x 10^-27 kg
Kinetic energy, K = 22 MeV
Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A
(a) time, t = 2.8 s
Let N be the alpha particles strike the surface.
N x 2e = q
N x 3.2 x 10^-19 = i t
N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8
N = 2.36 x 10^12
(b) Length, L = 16 cm = 0.16 m
Let N be the alpha particles
K = 0.5 x mv²
22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²
v² = 1.054 x 10^15
v = 3.25 x 10^7 m/s
So, N x 2e = i x t
N x 2e = i x L / v
N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)
N = 4153.85
(c) Us ethe conservation of energy
Kinetic energy = Potential energy
K = q x V
22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V
V = 1.17 x 10^7 V
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (0.314 kg) x (164 m/s²)
= 51.5 newtons
(about 11.6 pounds).
Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.
~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )
Answer:
5. -24 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s².
mathematically,
a = dv/dt ............................ Equation 1
Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.
But
v = dx(t)/dt
Where,
x(t) = 27t-4.0t³...................... Equation 2
Therefore, differentiating equation 2 with respect to time.
v = dx(t)/dt = 27-12t²............. Equation 3.
Also differentiating equation 3 with respect to time,
a = dv/dt = -24t
a = -24t .................... Equation 4
from the question,
At the end of 1.0 s,
a = -24(1)
a = -24 m/s².
Thus the acceleration = -24 m/s²
The right option is 5. -24 m/s²
Answer:
it will be d) 14.4W
Explanation:
potential difference (v) = 12 volts
resistance (r) = 10 ohms
now, we know
=》
=》
=》
=》
Explanation:
t = usin©/g
Where t is the time to reach the maximum height
Time spent in air is T = 2t
Hence, T = 2usin©/g
T = 2 x 20 x sin 65°/ 9.8
T = 3.69s
