2. Another airplane is flying in the opposite direction. It covers the same

A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric fric

gayaneshka [121]

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

$KE = \fract{1}{2}mv^2$

$PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))$

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

$m = 910 Kg$

$r_1 = 1200 + 6371 km = 7571km$

$r_2 = 6371 km,$

Replacing we have,

$\frac{1}{2} mv^2 = -GMm(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2GM(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})$

$v = 4456 m/s$

Therefore the speed of the object when striking the surface of earth is 4456 m/s

3 0

4 years ago

What is homeostasis?

SashulF [63]

<span>Homeostasis maintains internal balance of the cell.-- Things like shiver if cold sweating if hot to maintain internally </span>

3 0

4 years ago

Read 2 more answers

(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How

OLEGan [10]

Answer:

(a) Current flowing through truck battery is 180 A

(b) Time taken in calculator is 333.33 s

Explanation:

(a) Given:

The charge on the truck battery,q = 720 C

Time, t = 4.00 s

Consider I be the current flowing through truck battery.

The relation between current, charge and time is:

I = q/t

Substitute the suitable values in the above equation.

$I=\frac{720}{4}$

I = 180 A

(b) Given:

The charge on the calculator,q = 7.00 C

The current flowing through calculator, I = 0.3 mA = 0.3 x 10⁻³ A

Consider t be the time.

The relation between current, charge and time is:

t = q/I

Substitute the suitable values in the above equation.

$t=\frac{1}{0.3\times10^{-3} }$

I = 333.33 s

8 0

4 years ago

A sphere is filled with air. If the volume of the sphere is increasing at a rate of 569 cubic inches per minute, what is the rat

Veseljchak [2.6K]

Answer:565

Explanation:

6 0

3 years ago

Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it

makkiz [27]

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

Here, we have been given the height through which the ball drops;

Height of drop = 14.5m - 1.9m = 12.6m

The right motion equation is;

V² = U² + 2gh

V is the final velocity

U is the initial velocity = 0

g is the acceleration due to gravity = 9.8m/s²

h is the height

Now insert the parameters and solve;

V² = 0² + 2 x 9.8 x 12.6

V² = 246.96

V = √246.96 = 15.7m/s

4 0

3 years ago