Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Coulomb's law:
Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²
= (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C) / (1.0 m)²
= (8.99×10⁹ x 1×10⁻¹² / 1.0) N
= 8.99×10⁻³ N
= 0.00899 N repelling.
Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.
-- '4.0 kg masses'; don't need it.
Mass has no effect on the electric force between them.
-- 'frictionless table'; don't need it.
Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
<span>The component most affected by the collisions is vertical. The ball's vertical will either decrease or increase due to the collision. If the velocity is high during the collsion the ball's vertical will likely be higher and if the ball's velocity is low the vertical will be as well.</span>
Answer:
The answer of this question is :- Virtual image
