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3 years ago

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2. A liquid that is twice as dense as water is used in a barometer. With this barometer, atmospheric pressure would push the liq

uid to a height of 5.15m. Explain why.\

1 answer:

Scrat [10]3 years ago

8 0

Answer:

The reason why the height for the liquid is 5.18 m is because, for constant pressure, height is inversely proportional to density.

Explanation:

We know that pressure, P = ρgh where ρ = density of material, g = 9.8 m/s and h = height.

Since density of liquid = 2 × density of water = 2 × 1000 kg/m = 2000 kg/m³

Since atmospheric pressure = 101500 N/m²,

The height of liquid for this pressure is h = P/ρg = 101500/(2000 × 9.8) = 5.18 m

For the same pressure, the height for water is h = P/ρg = 101500/(1000 × 9.8) = 10.36 m

Since pressure = constant,

P = ρgh ⇒ P/ρg = h since P and g are constant, h ∝ 1/ρ

ρ₁h₁ = ρ₂h₂

So, the reason why the height for the liquid is 5.18 m is because, for constant pressure, height is inversely proportional to density.

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Answer:

True

Explanation:

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3 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

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3 years ago

A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a

wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

$\frac{V}{T}=const.$

Which can be also re-written as

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

where

$V_1, V_2$ are the initial and final volumes of the gas

$T_1, T_2$ are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

$V_1 = 7.00\cdot 10^2 mL = 700 mL$ is the initial volume

$T_1=20.0^{\circ}C+273=293 K$ is the initial absolute temperature

$V_2$ is the final volume

$T_2 = 1.00\cdot 10^2 K = 100 K$ is the final temperature

Solving for $V_2$ ,

$V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL$

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3 years ago

Write down short notes on:<br>c. Power

Fudgin [204]

Answer: In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.

Explanation: SI unit: watt (W)

In SI base units: kg⋅m2⋅s−3

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3 years ago

A tow truck is pulling 25,000 kg van with a tow hook that meets the van at an angle of 30° with the ground. How much force must

Law Incorporation [45]

Answer:

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Explanation:

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The force and the acceleration does not have the same direction in this case, so we need to decompose the force into its horizontal component, which is the force that will generate the horizontal acceleration:

Force_x = Force * cos(30)

Then, we have that:

Force_x = mass * acceleration

Force * cos(30) = 25000 * 3

Force * 0.866 = 75000

Force = 75000 / 0.866 = 86605.08 N

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3 years ago