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3 years ago

Physical quantity are measurable quantities

Physics

1 answer:

Vilka [71]3 years ago

7 0

Answer:

yeah physical quantities are the quantities which can be meaured

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Explanation:

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What is the speed of a bobsled whose distance-time graph indicates that it traveled 119m in 29s?

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3 years ago

Read 2 more answers

Q11) If you were standing at the top of a building and you dropped a rock.

Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

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2 years ago

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.

6 0

3 years ago