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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:

6 0

when 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 push 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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A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H

Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

$\theta =44.8{\circ}$

using $v^2-u^2=2as$

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration $(gsin\theta )$

s=distance traveled

$0-(1.72)^2=2(-9.81\times sin(44.8))s$

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

$0=1.72-gsin(44.8)\times t$

$t=\frac{1.72}{9.8\times sin(44.8)}$

t=0.249 s

(c)Speed of the block at bottom

$v^2-u^2=2as$

Here u=0 as it started coming downward

$v^2=2\times gsin(44.8)\times 0.214$

$v=\sqrt{2.985}$

v=1.72 m/s

3 0

3 years ago

Dwarf galaxies are generally only about one-fifth the size of the andromeda galaxy. true or false

Keith_Richards [23]

It is FALSE.........

5 0

3 years ago

Read 2 more answers

Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c

PIT_PIT [208]

Missing part in the text: "...the charges are separated by a distance of 30.0 cm."

Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
$E=k_e \frac{q}{r^2}$
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
$E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C$

While the electric field generated by the second charge is
$E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C$

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
$E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C$

4 0

3 years ago

A world-class tennis player can serve a tennis ball at 150 mi/h (about 67 m/s). The length of a tennis

Rudik [331]

Answer:

The expression for the time taken by an object to move with a speed at some distance is,

t= d/v

Here, t is the time taken by the opponent to react, d is the length of the court, andis the speed of the ball.

Explanation:

hope u like it :)

5 0

3 years ago

Explain why stars are unevenly distributed throughout the universe.

Norma-Jean [14]

Stars are not spread uniformly across the universe but are normally grouped into galaxies along with interstellar gas and dust. A typical galaxy contains hundreds of billions of stars and there are more than 2 trillion galaxies

8 0

2 years ago