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KIM [24]
2 years ago
10

A ball has 2 kg*m/s of momentum when thrown with a velocity of 8 m/s outwards. Find the mass of the ball.

Physics
1 answer:
worty [1.4K]2 years ago
5 0

Answer:

As, we know momentum = mass × velocity that is p = mv

we have momentum (p) = 2 kg m/s

and velocity (v) = 8 m/s

so substitute values in p=mv

2 kg m/s = m × 8 m / s

m = 2 / 8 kg = 0.25 kg

Therefore, mass of substance is 0.25 kg.

2/8 in fractional form.....

Hope it helps

Please mark me as the brainliest

Thank you

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A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon
larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_A

where,

p_A = partial pressure of substance A

p_T = total pressure

\chi_A = mole fraction of substance A

We are given:

m_{N_2}=2.80g

m_{Xe}=24.9g

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

And,

n_A=\frac{m_A}{M_A}

Mole fraction of nitrogen is given as:

\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}

Molar mass of N_2 = 28 g/mol

Molar mass of Xe =  g/mol

Putting values in above equation, we get:

\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}

\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345

To calculate the mole fraction of xenon, we use the equation:

\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655

p_{N_2}=836mmHg\times 0.345=288mmHg

p_{Xe}=836mmHg\times 0.655=548mmHg

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

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during a car ride Sandra travel led directly east for 20 miles and then directly south for 15 miles what was her displacement. 2
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