Answer:
B. the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.
Explanation:
To an observer at rest in the rocket who can't see either sides of the rocket, the speed of the light is constant which means the distance to the front or the back is same and would appear to reach the rocket at the same time.
Although from the point of view of the person on the earth, the front of the rocket is travelling in opposite direction of the light while the back of the rocket is moving closer to the light. This means that the distance travelled by the light going forward will be longer going backwards. And since the speed of light is constant in both directions, the light will reach the back of the rocket before it reaches the front for the observer on the earth.
A cloud in space that is composed of dust and gas is call a nebula. The word "nebula" is derived from the Latin word for "cloud," and nebulae in space are indeed large interstellar clouds made up of dust, hydrogen, helium and plasma.
Answer:
3.735×10⁻⁶ N
Explanation:
From newton' s law of universal gravitation,
F = Gmm'/r² .............................. Equation 1
Where F = Gravitational force between the person and the refrigerator, m = mass of the person, m' = mass of the refrigerator, r = distance between the person and the refrigerator. G = gravitational universal constant.
Given: m = 70 kg, m' = 200 kg, r = 0.5 m
Constant: G = 6.67×10⁻¹¹ Nm²/kg².
F = (6.67×10⁻¹¹×70×200)/0.5²
F = 93380×10⁻¹¹/0.25
F = 373520×10⁻¹¹
F = 3.735×10⁻⁶ N
Hence the force between the person and the refrigerator = 3.735×10⁻⁶ N
Answer: sheet of charge
Explanation:
a )
Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.
b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.
C ) Electric field which is almost uniform near the sheet of charge is equal t the following
E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²
E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²
= .3389 x 10³
= 338.9 V / m
spacing between 1 V
= 1 / 338.9 m
= 2.95 X 10⁻3 m
= 2.95 mm.