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KIM [24]
3 years ago
9

At which point does the planet have the least gravitational force acting on it?

Physics
2 answers:
Snezhnost [94]3 years ago
8 0

Answer:

Point A

Explanation:

Recall that gravitational force is inversely proportional to the square of the distance between the 2 objects.

This means that the farther away the objects are from each other, the lesser their gravitational interaction will be.

In this case we can see that Point A is furthest away from the star, hence this will be the location with the least gravitational force from the star.

Elza [17]3 years ago
5 0

Answer:

At which point does the planet have the least gravitational force acting on it?

Explanation:

In an elliptical orbit, when a planet is at its furthest point from the Sun, it is under the least amount of gravity, meaning that the force of gravity is strongest when it is closest.

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The following letter are used to help remember the characteristics of minerals:S I C C N. What does the S stand for?
Dennis_Churaev [7]

Your answer would be A. Solid. Hope this helps!

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State the condition for maximum current to be drawn from the cell
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Use the equation I=V/R where I is current and V is the voltage plus R is the resistance so when voltage is the highest and resistance is lowest the current is the highest
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Uma wants to recreate the discovery of superconductivity. Which describes the materials needed?
zaharov [31]

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2 years ago
Read 2 more answers
the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,
balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0
3 years ago
A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl
guajiro [1.7K]

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a  = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b  = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

P_1 = m_av_a + m_b v_b

P_1 = 1300 \times 35+ 1000 \times 25

P_1 =70500 Kg.m/s

final momentum

P_2 = m_aV_a + m_b V_b

P_2 = 1300 \times 30+ 1000 \times 31.5

P_2 =70500 Kg.m/s

here  initial momentum is equal to the final momentum of the car.

hence, momentum is conserved in the collision.

6 0
3 years ago
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