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3 years ago

A 62 kg skydiver moving at terminal speed falls 50 m in 1 s. What power is the skydiver expending on the air?

Physics

1 answer:

IgorLugansk [536]3 years ago

3 0

Answer:

P = 30380 W

Explanation:

given,

mass of the skydiver, m = 62 Kg

distance, s = 50 m

time, t = 1 s

we know,

$Power= \dfrac{work\ done}{time}$

$Power= \dfrac{F.s}{t}$

F = m g

$Power= \dfrac{m g.s}{t}$

inserting all the values

$P = \dfrac{62\times 9.8\times 50}{1}$

P = 30380 W

hence, Power the skydiver expending on the air is 30380 W

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True or False: Every lightning strike produces thunder.

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False,

True,

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Answer:

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3 years ago

He shoots the tree stump, which has a mass of (M), with a bullet of mass (m) traveling at some velocity (vbullet), and the bulle

Ymorist [56]

Answer:

Explanation:

Given

mass of tree stump is $M$

mass bullet is $m$

velocity of bullet is $v$

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4 0

2 years ago

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33Ã—Ã—10âˆ

AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, $F=3.33\times 10^{-21}\ N$

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

$F=k\dfrac{q^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}$

$q=1.21\times 10^{-16}\ C$

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

$n=\dfrac{q}{e}$

$n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}$

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0

3 years ago

A car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collisio

Sergeeva-Olga [200]

Answer:

Explanation:

The vehicle is experiencing a large force created by the concrete wall.

Equation

vf^2 = vi^2 + 2*a * d

Givens

vf = 0 The car eventually does stop.

vi = 72 km/hr * [ 1000 m/ km] * [1 hour / 3600 seconds]

vi = 20 meters / second

a = ?

m = 850 kg

Solution

vf^2 = vi^2 + 2a*d

0 = 20 m/s + 2* 2 *a

-20 m/s = 4a

-20/4 = a

a = - 5 m/s^2 The minus sign tells you the vehicle is slowing down. It sure should be.

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F = - 850 * (-5)

F = - 4250 N

The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.

8 0

3 years ago