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3 years ago

how do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate

1 answer:

3 0

While skydiving, its not just freely falling under Earth's gravity. Additional force called drag acts against the gravity which slows down the rate of fall. Drag is caused by the air molecules which pushes against the body as it falls through them. This is actually a significant amount of force which slows down the rate of fall of the body. Drag depends on the contact surface area and weight. 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 because of the less contact surface area of the body with the air molecules while in the former case. No two persons have identical body shape and weight. Hence, 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.

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3 years ago

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4 years ago

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13. Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distanc

Gennadij [26K]

This question is incomplete, the complete question is;

Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).

how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers

Answer:

Given the data in question;

Dipole moment P = 1 × 10⁻⁹ C.m

now dipole pointing to the right;

P→

$_{-\theta }$ (-) ---------------->(+) $_{+\theta }$

so let distance between the dipoles be d

∴ P = d $\Theta$

Let $\Theta_{1}$ = 1 nC

P = d $\Theta$

1 × 10⁻⁹ = 1 × 10⁻⁹ × d

d = (1 × 10⁻⁹) / (1 × 10⁻⁹)

d = 1 m

Also Let $\Theta_{2}$ = 2 nC

P = d $\Theta$

1 × 10⁻⁹ = 2 × 10⁻⁹ × d

d = (1 × 10⁻⁹) / (2 × 10⁻⁹)

d = 0.5 m

Also Let $\Theta_{3}$ = 3 nC

P = d $\Theta$

1 × 10⁻⁹ = 3 × 10⁻⁹ × d

d = (1 × 10⁻⁹) / (3 × 10⁻⁹)

d = 0.33 m

such that;

charge distance

1 nC 1.00 m

2 nC 0.50 m

3 nc 0.33 m

4 nC 0.25 m

5 nC 0.20 m

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3 years ago