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Lady bird [3.3K]

3 years ago

9

What should a free-body diagram look like for a skydiver who has opened his parachute and is now slowing down as he falls?

2 answers:

iogann1982 [59]3 years ago

7 0

answer A is correct i took the test

Ksivusya [100]3 years ago

5 0

Answer:

There is an arrow up for air resistance and an arrow down for gravity. The arrow up is longer than the arrow down.

Explanation:

The text of the problem says that the skydiver is slowing down: this means that he has an acceleration, which is directed opposite to the motion of the skydiver. Since the motion is downward, the acceleration must be upward.

There are two forces acting on the skydiver: the gravity (downward) and the air resistance (upward). According to Newton's second law:

F=ma

the acceleration has the same direction of the net force, so the net force must also be upward: therefore, the air resistance must be greater than the gravity, so the arrow up for air resistance is longer than the arrow down for gravity.

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Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .

Vlad [161]

<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

Work done = Force × distance

Force = 95 Newtons

Distance = X2 -X1

= 4 - (-5)

= 9 cm

Thus;

work done = 95 × 9/100

<u>= 8.55 Joules </u>

5 0

3 years ago

es la atmósfera que se percibe en un cuento de acuerdo a la manera de pensar y sentir de los personajes no se ve sólo se siente

masya89 [10]

Answer:

can you please ask in english

8 0

3 years ago

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place

anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

4 0

3 years ago

Deperate need of HELP! MARKING BRAINLIEST

Shtirlitz [24]

Answer:

yes, what's your question. l will help if I can.

6 0

3 years ago

Read 2 more answers

rosijanka [135]

Answer:

its deepend on where you stand but. most cases g = 10 m/s²

W = M X G

W = 13 kg x 10 m/s²

W = 130 N

Explanation:

option b is correct brainlist ke

3 0

2 years ago