Assume it is in uniform deceleration.
u=25
v=0
s=x
a=?
t=12
v=u+at
0=25+a(12)
a=-2.08(3sig fig)
decelerarion of 2.08 ms^(-2)/
acceleration=-2.08ms^(-2)
Answer:
(a) 272.73 m
(b) 0.338 N/C
Explanation:
frequency, f = 1100 kHz = 1100 x 1000 Hz
E(t) = Eo Sin(2πft)
Eo = 0.62 N/C
(a) Velocity of light, c = 3 x 10^8 m/s
wavelength, λ = c / f = (3 x 10^8) / (1100000) = 272.73 m
Thus, the wavelength is 272.73 m.
(b) at t = 3.1 microsecond = 3.1 x 10^-6 s
E = Eo Sin (2 π ft)
E = 0.62 Sin (2 x 3.14 x 1100 x 10^3 x 3.1 x 10^-6)
E = 0.62 Sin (21.4148)
E = 0.62 x 0.5449 = 0.338 N/C
Thus, the electric field at t = 3.1 microsecond s 0.338 N/C.
Answer:
I will answer in English.
Here we will use the relation
Velocity*time = distance
So:
a) velocity = 3m/s
time = 2s
Distance = 3m/s*2s = 6m
b) velocity = 2m/s
time = 3.5s
Distance = 2m/s*3.5s = 7m
c) velocity = 10m/s
time = 0.5s
Distance = 10m/s*0.5s = 5m
d) velocity = 4m/s
time = 2.5s
Distance = 4m/s*2.5s = 9m
e) velocity = 1.5m/s
time = 5s
Distance = 1.5m/s*5s = 7.5m
Answer:
8W
Explanation:
Given parameters:
Distance covered = 20m
Work done = 120J
Time = 15s
Unknown:
How much power does the Ox exerts = ?
Solution:
Power is the rate at which work is being done
Power =
Power exerted by ox =
= 8W