Answer:
Since incident wave and its reflected part in opposite phase superimpose on each other
So correct answer will be
Option B
Explanation:
Here we know that the wave reflection is done by rigid boundary
So when wave is reflected by the boundary then its phase is reversed by 180 degree
so the reflected wave is in reverse phase from the boundary
so we can superimpose the reflected part with incident wave to dine the resultant wave
So the phenomenon is given as follow
This is just testing your ability to recall that kinetic energy is given by:
<span>k.e. = ½mv² </span>
<span>where m is the mass and v is the velocity of the particle. </span>
<span>The frequency of the light is redundant information. </span>
<span>Here, you are given m = 9.1 * 10^-31 kg and v = 7.00 * 10^5 m/s. </span>
<span>Just plug in the values: </span>
<span>k.e. = ½ * 9.1 * 10^-31 * (7.00 * 10^5)² </span>
<span>k.e. = 2.23 * 10^-19 J
so it will be d:2.2*10^-19 J</span>
Answer:
The magnitude of the induced voltage in the loop is 20 mV.
Explanation:
given;
length of loop, L = 0.43 m
width of loop,w = 0.43 m
velocity of moved loop, v = 0.15m/s
magnetic field strength,B = 0.31 T
To determine the magnitude of the induced voltage in the loop, we apply Faraday's law;
magnitude induced E.M.F = BLv
magnitude induced E.M.F = 0.31 x 0.43 x 0.15 = 0.02 V = 20 mV
Therefore, the magnitude of the induced voltage in the loop is 20 mV.
Answer:
a) X = 17.64 m
b) X = 17.64 + 4∆t^2 + 16.8∆t
c) Velocity = lim(∆t→0)〖∆X/∆t〗 = 16.8 m/s
Explanation:
a) The position at t = 2.10s is:
X = 4t^2
X = 4(2.10)^2
X = 17.64 m
b) The position at t = 2.10 + ∆t s will be:
X = 4(2.10 + ∆t)^2
X = 17.64 + 4∆t^2 + 16.8∆t m
c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,
∆X= 4∆t^2 + 16.8∆t
Divide by ∆t on both sides:
∆X/∆t = 4∆t + 16.8
Taking the limit as ∆t approaches to zero we get:
Velocity =lim(∆t→0)〖∆X/∆t〗 = 4(0) + 16.8
Velocity = 16.8 m/s
Answer:
Explanation:
let force exerted by engine be F.Net force =( F-400)N, applying newton law
F-400 = 1.5 x 10³x18 =27000 ,
F = 27400 N.
velocity after 12 s = 0 + 18 x 12 = 216 m/s
Average velocity = (0 + 216 )/2 = 108 m/s
Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶
b) At 12 s , velocity = 216 m/s
Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.
