Answer:
3.64×10⁸ m
3.34×10⁻³ m/s²
Explanation:
Let's define some variables:
M₁ = mass of the Earth
r₁ = r = distance from the Earth's center
M₂ = mass of the moon
r₂ = d − r = distance from the moon's center
d = distance between the Earth and the moon
When the gravitational fields become equal:
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
M₁ / r² = M₂ / (d − r)²
M₁ / r² = M₂ / (d² − 2dr + r²)
M₁ (d² − 2dr + r²) = M₂ r²
M₁d² − 2dM₁ r + M₁ r² = M₂ r²
M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0
d² − 2d r + (1 − M₂/M₁) r² = 0
Solving with quadratic formula:
r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)
When we plug in the values, we get:
r = 3.64×10⁸ m
If the moon wasn't there, the acceleration due to Earth's gravity would be:
g = GM / r²
g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²
g = 3.34×10⁻³ m/s²
Answer:
the action of investigating something or someone; formal or systematic examination or research.
Explanation:
This definition is provided by Oxford Languages
Answer:
no it would not. that is an open circuit and it would need to be closed at the switch for current to flow.
Answer:
ΔP = 14.5 Ns
I = 14.5 Ns
ΔF = 5.8 x 10³ N = 5.8 KN
Explanation:
The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:
ΔP = mv₂ - mv₁
ΔP = m(v₂ - v₁)
where,
ΔP = Change in Momentum = ?
m = mass of ball = 0.145 kg
v₂ = velocity of batted ball = 55.5 m/s
v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)
Therefore,
ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)
<u>ΔP = 14.5 Ns</u>
The impulse applied to a body is equal to the change in its momentum. Therefore,
Impulse = I = ΔP
<u>I = 14.5 Ns</u>
the average force can be found as:
I = ΔF*t
ΔF = I/t
where,
ΔF = Average Force = ?
t = time of contact = 2.5 ms = 2.5 x 10⁻³ s
Therefore,
ΔF = 14.5 N.s/(2.5 x 10⁻³ s)
<u>ΔF = 5.8 x 10³ N = 5.8 KN</u>
Answer:
That the polar air has has more pressure than the air at the equator.
Explanation:
