3 years ago

Please hurry!!!

1 answer:

3 0

First we need to know the equation which is P=mv
Use the formula- 3.2 kg x m/s = 762 x V
divide 762 in both side to get the v by itself
now we will have 0.004193971166 and lets round it!
then we will have the answer of .0042

You might be interested in

Si un vector tiene una dirección de 2300 a partir del eje x positivo, ¿Qué signos tendrán sus componentes x y y? Si la la razón

Masteriza [31]

If you can write this in English I can help

8 0

3 years ago

A softball has a mass of 0.18 kg. What is the gravitational force on Earth due to the ball, and what is the Earth's resulting ac

alexgriva [62]

Answer:F=1.7802

Explanation:

Since we've been given the mass to be .18kg,we are asked to find the Force of which the formulae is

F=ma where f-force,m-mass and a-acceleration due to gravity

So we can just substitute

F-?.m=.18 and a9.89

F=.18×9.89

F=1.7802N

6 0

2 years ago

Which element is the main component of<br> steel and the most widely used of all metals

Blizzard [7]

Answer:

iron

Explanation:

8 0

2 years ago

Read 2 more answers

You observe a star cluster with a main-sequence turn-off point at spectral type G2 (the same spectral type as the Sun). What is

Goryan [66]

Answer i dont even know im just putting this cus id ont care

Explanation:

3 0

2 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago