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3 years ago

11

A skydiver jumps out of a plane flying at an altitude of 5,500 meters. How fast is the skydiver going when they get to the groun

d ?

1 answer:

andrey2020 [161]3 years ago

4 0

Answer:

2,500 feet (760 meters)

Explannation: <em>At about 2,500 feet (760 meters), the skydiver throws out a pilot chute, and it deploys the parachute. Its used to control the fall rate.</em>

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Though some self-report error may exist, surveys/interviews are useful in that they allow researchers to collect large amounts o

Leokris [45]

Definitely true, surveys and interviews aren’t flawless but you can collect lots of data from them

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3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

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3 years ago

What type of planet is Venus?

mafiozo [28]

Venus would be an inner planet!
x

7 0

3 years ago

Read 2 more answers

Real bodies emit and absorb more radiation than a blackbody at the same temperature. True or False

monitta

Answer: False

Explanation:

Relative to the concept of radiations, a black body is an object capable of absorbing any form of electromagnetic radiation irrespective of its frequency or angle of incidence when incident on such object.

However, the same cannot be said about real bodies as real bodies are those which reflect all rays incident on them completely and uniformly in all directions.

One very important characteristic of black bodies is that they are ideal emmiters.

The concept of emmisivity is brought about by the existence of real bodies .

This is due to the fact that they are only able to emit radiation at a fraction of the black body energy levels.

Please note that by convention, the emmisivity of a real body is always less thaan 1.

As such they are not able to emit as much radiation as a black body at the same temperature.

3 0

3 years ago