Answer:
a) According to Newton's law of gravitation, as the distance between the Moon and the Earth decreases, the gravitational attraction increases and vice versa
The gravitational force of the Moon on the Earth causes the Earth to be slightly bulged on the side directly facing the Moon
The gravitational force also pulls the water bodies on the Earth's surface towards the Moon in the same manner and the effect is more pronounced due to the ability of the liquid water to assume a shape based on the magnitude of the gravitational field attracting it
Therefore, the region where the Moon is closest to the Earth we have a high tide as the water level rises and the region which is perpendicular to where the Moon is located has a low tide
b) The two special types of tides are
1) The neap tide
2) The spring tide
Neap tide
Neap tide occurs when the Sun and Moon are 90° apart from each other when they are viewed by an observer from Earth
The gravitational pull of the Sun cancels (partially) the effect of the gravitational pull and tidal force of the Moon, resulting in minimum tidal range
Spring Tide
Spring tide occurs when the Earth, the Moon, and the Sun are simultaneously inline, such that the Sun reinforces the gravitational pull and tidal force of the Moon, resulting in a maximum tidal range
Explanation:
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
Answer:
C.
Measure the wear on his treads before and after riding a certain number of laps.
Answer:
2 s, -20 m/s
Explanation:
Given:
y₀ = 20 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t and v
y = y₀ + v₀ t + ½ at²
0 = 20 + 0 + ½ (-9.8) t²
0 = 20 − 4.9 t²
t ≈ 2 s
v² = v₀² + 2a(y − y₀)
v² = 0 + 2(-9.8)(0 − 20)
v ≈ ±20 m/s
Since the rock is falling, v = -20 m/s.
Explanation:
The temperature in the inner solar system was too high for light gases to condense, while in the outer solar system, the temperature was much lower, which allowed the Jovian planets to form, which grew enough to accumulate and retain the hydrogen gas that remained in the solar nebula, which led to its high levels of hydrogen and large size.
