Sky diving involves free fall under gravity along with the drag due to air molecules pushing against the body slowing the rate of fall of a body. This is actually a significant amount of force. The drag force depends on the contact surface area and weight of the body. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position. This is because of the less contact surface area of the body with the air molecules while in the former case. Since no two persons have identical body shape and weight, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.
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>
- Mass of the elevator (m) = 570 Kg
- Acceleration = 1.5 m/s^2
- Distance (s) = 13 m
- Let the force be F.
- We know, F = ma,
- Therefore, F = (570 × 1.5) N = 855 N
- Angle between distance and force (θ) = 0°
- We know, work done = F s Cos θ
- Therefore, work done by the cable during this part
- = (855 × 13 × Cos 0°) J
- = (855 × 13 × 1) J
- = 11115 J
<u>Answer</u><u>:</u>
<u>1</u><u>1</u><u>1</u><u>1</u><u>5</u><u> </u><u>J</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Explanation:
o has Celsius of indirect measurment
I would say true. If you are calculating using vectors than it would need both...
