A commercial diffraction grating has 300 lines per mm. When a student shines a 470 nm laser through this grating, how many brigh
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The correct answer is "As the distance from the earth increases, the gravitational pull on the spaceship would decrease."
In fact, the gravitational force (attractive) exerted by the Earth on the spaceship is given by
where G is the gravitational constant, M the Earth's mass, m the mass of the spaceship and d the distance of the spaceship from the Earth. As we can see from the formula, as the distance d between the spaceship and the Earth increases, the gravitational force F decreases, so answer D) is the correct one.
In fact, the gravitational force (attractive) exerted by the Earth on the spaceship is given by
where G is the gravitational constant, M the Earth's mass, m the mass of the spaceship and d the distance of the spaceship from the Earth. As we can see from the formula, as the distance d between the spaceship and the Earth increases, the gravitational force F decreases, so answer D) is the correct one.
Answer: 1.51 km
Explanation:
<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.
Or,
Where Q1 and Q2 are magnitude of two charges and r is distance between them:
<u>Given:</u>
Q1 = Charge near top of cloud = 48.8 C
Q2 = Charge near the bottom of cloud = -41.7 C
Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N
k = 8.99 x 109Nm^2/C^2
<u>So,</u>
Therefore, the separation between the two charges (r) = 1.51 km