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prohojiy [21]
3 years ago
15

Is air made of matter ?​evidence and reasoning.

Physics
1 answer:
Fiesta28 [93]3 years ago
4 0
Yes, air is matter because it is a gas composed mostly of nitrogen, oxygen, and argon, and gases do have mass because they’re molecules bouncing around a contained volume (here, the atmosphere)
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a 46 kilogram student climbs 11 meter up a rope at a constant speed if the student power output is 230 watts how long in seconds
Bezzdna [24]
230×46=10580÷11=961 second
3 0
3 years ago
Why can't a hot air balloon go to higher layers?​
Elena L [17]
If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.
5 0
3 years ago
One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e
frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is 1.122 X 10^{-7}  kg

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

E= \Delta mc^{2}

where \Delta m = change in mass

c = speed of light = 3 \times 10 ^{8}m/s

Making m the subject of the formula, we can find the change in mass to be

\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0
3 years ago
Every winter I fly to Michigan. It's a total distance of 3900km. It takes 5 hours. What is my average speed?
olga_2 [115]
\frac{3900}{5} = 780\sf ~kilometer~per~hour

Your average speed is 780 kilometer per hour.
4 0
3 years ago
Read 2 more answers
A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

3 0
3 years ago
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