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nevsk [136]
2 years ago
8

H2+o2 - h2o mass Please help

Physics
1 answer:
emmasim [6.3K]2 years ago
4 0

H2O2 H2O

H2 + O2 → H2O. Word equation: Hydrogen gas + Oxygen gas → water. Type of Chemical Reaction: For this reaction we have a Combination reaction. Balancing Strategies: For this reaction it is helpful to start by changing the coefficient in front of H2O and so that you have an even number of oxygen atoms.

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9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.
SVETLANKA909090 [29]

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

W = E_p = mgh

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

W = 50*9.8*13 = 6370 J

3 0
2 years ago
What is the temperature 32 degrees Fahrenheit in degrees Celsius? A. 0°C B. 20°C C. 10°C D. –10°C
omeli [17]
The answer would 0. The reasoning of this is because freezing point in celsius is always 0 degrees but in fahrenheit the freezing point is 32 degrees.
8 0
2 years ago
Which change will cause the gravitational force between a baseball and a soccer ball to increase ? Check all that apply
Alex73 [517]
Moving the balls closer together pumping more air into the soccer ball putting the baseball in a plastic case 
4 0
3 years ago
Read 2 more answers
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
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Olin [163]

Answer:

c

Explanation:

i did the test

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2 years ago
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