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3 years ago

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is J.

Physics

2 answers:

Svetlanka [38]3 years ago

8 0

2.5 joules according to the formula k=1/2MVsquare we would get 2.5 joules as the answer

DENIUS [597]3 years ago

5 0

25 joules would be the correct answer

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Explain the differences in structure and use for life between oxygen gas in the atmosphere and ozone.

navik [9.2K]

Answer:

Most of the oxygen in the atmosphere is in the form of O2, which means it is made up of molecules containing two oxygen atoms. Ozone, however is O3, which means it is made up of molecules containing three oxygen atoms. O2 is what we breath, and what plants release from photosynthesis. Ozone occurs naturally high in the stratosphere, where it absorbs ultraviolet light, protecting us here on the surface from skin cancer. Ozone can also occur closer to the surface of the earth as a pollutant. It is formed from reactions with O2 and chemicals emitted from factories and cars. It comes in the form of smog.

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2 years ago

PLEASE HELP MEEEEEE!!!!!!!!!!!!!!!!!!

Simora [160]

You would need to freeze it in a freezer. Hope this helps if it does could I have brainlist thanks

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2 years ago

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A 1kg mass is thrown to a height of 2cm. what is the potential energy

prohojiy [21]

Mass=m=1kg
Height=h=2cm=0.02m
Acceleration due to gravity=g=10m/s^2

$\\ \tt\hookrightarrow P.E=mgh$

$\\ \tt\hookrightarrow PE=1(10)(0.02)$

$\\ \tt\hookrightarrow PE=0.2J$

4 0

1 year ago

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In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t

galina1969 [7]

Answer:

$W_x = 0.9156\ c$

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

$W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}$

u_x = 0.543 c

v_x = - 0.741 c

$W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}$

$W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}$

$W_x = \dfrac{1.284c}{1+0.402363}$

$W_x = 0.9156\ c$

Relative velocity of the particle is $W_x = 0.9156\ c$

5 0

2 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km.A)If you and your spacesuit ha

WITCHER [35]

A) 0.189 N

The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

$F=\frac{GMm}{R^2}$

where

G is the gravitational constant

8.7×10^13 kg is the mass of the asteroid

m = 130 kg is the mass of the man

R = 2.0 km = 2000 m is the radius of the asteroid

Substituting into the equation, we find

$F=\frac{(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)(130 kg)}{(2000 m)^2}0.189 N=$

B) 2.41 m/s

In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

$\frac{GMm}{R^2}=\frac{mv^2}{R}$

where

v is the speed of the astronaut

Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)}{2000 m}}=2.41 m/s$

3 0

3 years ago