Electric field due to a charged rod along its axis is given by

here we know that
L = 14 cm
r = distance from end of rod
r = 36 - 7 = 29 cm
Q = 222 mC
now we will have


<h2>Explanation:</h2><h3>3. </h3>
When light bounces back, it is <em>reflected</em>. (That's why you see your <em>reflection</em> in a mirror.) When light is bent from the path it is taking, it is <em>refracted</em>. The only answer choice that makes correct use of these terms is the third choice:
- Part of the ray is <em>refracted</em> into ray B; part of the ray is <em>reflected</em> as ray R.
_____
<h3>4.</h3>
The index of refraction is the ratio of the sine of the angle of incidence to the sine of the angle of refraction. Both angles are measured from the normal to the surface. The angle of refraction here is 12.5° less than the angle of incidence, 44°, so is 31.5°. Then the index of refraction of the medium is ...
n = sin(44°)/sin(31.5°) = 0.69466/0.52250 = 1.3299 ≈ 1.33
- none of the offered choices is correct. The closest is 1.34.
<u>Answer:</u>
<em>The amount of water entering the earth through precipitation is equal to the amount of water leaving earth through transpiration.</em>
<u>Explanation:</u>
Rates of precipitation and evaporation vary widely according to regions and seasons. But in a global scale the rates are equal. Thus the total amount of earth’s water maintains its constancy even though there is a continuous change in forms of water.
Evaporation and transpiration are the forms in which Water leaves the earth and it returns to the earth in various forms of precipitation like rain, snow, dew, fog etc. This water then reaches ocean and land. The water that reaches the land flows as surface run off into rivers and water bodies or seep into the ground replenishing the ground water table.
Answer:
Explanation:
mass of probe m = 474 Kg
initial speed u = 275 m /s
force acting on it F = 5.6 x 10⁻² N
displacement s = 2.42 x 10⁹ m
A )
initial kinetic energy = 1/2 m u² , m is mass of probe.
= .5 x 474 x 275²
= 17923125 J
B )
work done by engine
= force x displacement
= 5.6 x 10⁻² x 2.42 x 10⁹
= 13.55 x 10⁷ J
C ) Final kinetic energy
= Initial K E + work done by force on it
= 17923125 +13.55 x 10⁷
= 1.79 x 10⁷ + 13.55 x 10⁷
= 15.34 x 10⁷ J
D ) If v be its velocity
1/2 m v² = 15.34 x 10⁷
1/2 x 474 x v² = 15.34 x 10⁷
v² = 64.72 x 10⁴
v = 8.04 x 10² m /s
= 804 m /s