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tatyana61 [14]
3 years ago
12

What is the momentum of a 37-kg person riding south on an 18-kg bicycle at 1.2 m/s?

Chemistry
1 answer:
guajiro [1.7K]3 years ago
5 0
P = m*v

p = momentum
m = mass
v = velocity

p = (37 kg + 18 kg)*(1.2 m/s)

p = (55 kg)*(1.2 m/s)

p = 66 kg-m / s

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What is the mass in grams of 3x10 atoms of helium
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Answer:

here

Explanation:

What is the total mass of $$3.01 x 10^23$$ atoms of helium gas?

✓ Well $$"Avogadro's number"$$ of helium atoms has a mass of $$4.0*g$$. Explanation: And $$"Avogadro's number"$$

3 0
3 years ago
What happens to gas particles when they are compressed?
Anuta_ua [19.1K]

Answer:

3)Some gas molecules move further apart and some move closer together.

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because When more gas particles enter a container, there is less space for the particles to spread out, and they become compressed. The particles exert more force on the interior volume of the container. This force is called pressure. There are several units used to express pressure.

4 0
2 years ago
- How much power does it take to lift a<br> 1,000 N load 10 m in 20 s?
Mariana [72]

Answer:

"500 Joule/sec" is the right answer.

Explanation:

The given values are:

Force,

F = 1000 N

Velocity,

s = 10 m

Time,

t = 20 s

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=  \frac{Force\times Velocity}{Time}

On putting the values, we get

=  \frac{1000\times 10}{20}

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5 0
3 years ago
Hydrogen is a gas at a room temperature, but it is found with the alkali metals in group 1 (1A) on the periodic table. Which is
poizon [28]
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4 0
3 years ago
What is the enthalpy of combustion (per mole) of C4H10 (g)? 
Artemon [7]
The balanced chemical reaction for the complete combustion of C4H10 is shown below:

                    C4H10 + (3/2)O2 --> 4CO2 + 5H2O

The enthalpy of formation are listed below:
          C4H10: -2876.9 kJ/mol
              O2:   none (because it is pure substance)
             CO2: -393.5 kJ/mol
             H2O: -285.8 kJ/mol

The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.

               ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
                       = -<em>126.1 kJ</em>

Thus, the enthalpy of combustion of the carbon is -126.1 kJ. 
5 0
2 years ago
Read 2 more answers
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