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nataly862011 [7]
3 years ago
11

If you ran 15 km/h for 20 min, how much distance would you cover?

Physics
1 answer:
nikitadnepr [17]3 years ago
4 0
Since we have 15 kilometers per hour, and we're looking for 20 minutes, let's set up proportions.
20/60 minutes = x/15
20/60 = 1/3, so let's leave that simplified.
1/3 = x/15
Look at the denominators, 3 to 15 is a factor of 5, so multiply the numerator by 5.
1 • 5 = 5, so you will cover 5 kilometers in 20 minutes.

I hope this helps!
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A sample of nitrogen occupies 5.50 liters under a pressure of 900 torr at 25oC. At what temperature will it occupy 10.0 liters a
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Answer:

<u>At 268.82°C</u> volume occupied by nitrogen is 10 liters at pressure of 900 torr.

Explanation:

Given:

Volume of a sample of nitrogen = 5.50 liters

Pressure = 900 torr

Temperature = 25°C

To find the temperature at which the nitrogen will occupy 10 liters volume at same pressure.

Solution:

Since the pressure is kept constant, so we can apply the temperature-volume law also called the Charles Law.

Charles Law states that the volume of a gas held at constant pressure is directly proportional to the temperature of the gas in Kelvin.

Thus, we have :

V ∝ T

\frac{V}{T}=k

where k is a constant.

For two samples of gases, the law can be given as:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

From the data given:

V_1=5.5\ l

T_1=25\ \°C =(273+25)K= 298 K

V_2=10\ l

We need to find T_2.

Plugging in values in the formula.

\frac{5.5}{298}=\frac{10}{T_2}

Multiplying both sides by T_2.

T_2\times\frac{5.5}{298}=\frac{10}{T_2}\times T_2

\frac{5.5}{298}T_2={10}

Multiplying both sides by \frac{298}{5.5}

\frac{298}{5.5}\times\frac{5.5}{298}T_2=\frac{10\times 298}{5.5}

T_2=541.82\ K

T_2=541.82\ K-273\ K = 268.82\°C

Thus, at 268.82°C volume occupied by nitrogen is 10 liters at pressure of 900 torr.

7 0
3 years ago
A 105 kg astronaut carrying a 16 kg tool bag finds himself separated from his spaceship by 18 m and moving away from the spacesh
Ratling [72]

Answer:

T=22.5sec

Explanation:

From the question we are told that:

Mass of astronaut m_a=105kg

Mass of tool m_t=16kg

Distance d=18m

Velocity of separation v_s= 0.1m/s

Velocity of tool bag v_t=4.5m/s

Generally the equation for momentum is mathematically given by

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Therefore

Initial Momentum before drop

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Initial Momentum after drop

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Therefore

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Generally the equation for Time T is mathematically given by

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 T=22.5sec

3 0
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