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nika2105 [10]
3 years ago
5

A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the ma

nometer was 12.2 cm. Atmospheric pressure was 783 mmHg. The pressure of the argon in the container was __________ mmHg.
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

122 mmHg

Explanation:

In the given problem, the height of the column will not be affected by the atmospheric pressure because the manometer end is closed.

The difference of height of the mercury level is given as,

h=12.2cm\\h=12.2(\frac{10mm}{1cm})\\ h=122mm

Now as we know that the pressure of the gas is equal to the height in the case of closed end manometer.

P_{gas}=h

Therefore,

P_{gas}=122 mmHg

This pressure is caused due to the presence of gas.

Therefore, the pressure of the argon gas in the container is 122 mmHg.

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A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The
nordsb [41]

Answer:

t=6.96s

Explanation:

From this exercise, our knowable variables are <u>hight and initial velocity </u>

v_{oy}=96ft/s

y_{o}=112ft

To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}

Solving for t using quadratic formula

t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}

a=-\frac{1}{2} (32.2)\\b=96\\c=112

t=-0.999s or t=6.96s

<u><em>Since time can't be negative the answer is t=6.96s</em></u>

7 0
3 years ago
Hii:) I need help with qn 1 &amp; please explain if possible too :) , thanks!
andrezito [222]

1. In the first 1.5 seconds, the lift accelerates from 0 to 3m/s. By definition, the acceleration is the ratio between the change in velocity and the time elapsed to change the velocity.

The change in velocity is \Delta v = v_{\text{final}}-v_{\text{initial}}=3-0=3.

The time elapsed is 1.5 seconds, so the acceleration is

a = \dfrac{3}{1.5}=2

meters per second squared.

2. We know, from the previous point, that the lift travelled 20m from the first floor. Since it returns to the first floor after the ascent, it must travel again those same 20m, just in reverse (descending instead of ascending). So, the total distance travelled is 20+20=40 meters.

The displacement, though, is zero, because it measures the distance between the starting and ending point of a certain motion. Since the lift starts and ends its motion at the same place (the first floor), its total displacement is zero.

3 0
3 years ago
Q4. Consider the skier on a slope shown in the figure below. Her mass including equipment is 55.0 kg.
Shkiper50 [21]

Answer:

Part a)

When there is no friction then acceleration is

a = 4.14 m/s^2

Part b)

if there is friction force along the inclined then acceleration is

a = 3.33 m/s^2

Explanation:

Part a)

As we know that the skier is on inclined plane

So here if there is no friction then net force along the inclined plane is given as

F = mg sin\theta

now acceleration of the skier is given as

a = \frac{F}{m}

a = g sin\theta

a = 9.81(sin25)

a = 4.14 m/s^2

Part b)

if there is friction force along the inclined then net force along the inclined plane is given as

F = mg sin\theta - F_f

now acceleration of the skier is given as

a = \frac{F}{m}

a = g sin\theta - \frac{F_f}{m}

a = 9.81(sin25) - \frac{45}{55}

a = 3.33 m/s^2

5 0
4 years ago
Explain why a clothes iron with a metal base must be connected to the mains by a three-core cable.
Lapatulllka [165]
The charge going to the clothes iron needs to be grounded so the iron does not make any shock
3 0
3 years ago
The molar weight of water vapor is about 18g per mole, while the molar weight of dry air is about 29g per mole since almost all
musickatia [10]

Answer:

The moist air mass would be denser

Explanation:

Density is defined as mass per unit volume. Hence the density of a substance (solid, liquid or gas) is directly proportional to its mass and inversely proportional to the volume occupied.

The mass of a gas is the product of its number of moles and its molar mass (mass = number of moles × molar mass), which indicates that the mass is directly proportional to molar mass, so the higher the molar mass, the higher the mass of different gases at equal volumes, temperature and pressure.

From the information given, the molar weight of dry air = 29g/mole.

The molar weight of moist air = molar weight of dry air + molar weight of water vapour = 29 + 18 = 47g/mole.

Therefore since higher molar mass transits to higher mass, it can be said that moist air of molar mass 47g/mole is denser than dry air of molar mass 29g/mole at equal volume, temperature and pressure.

Simple picture the two gasses in two transparent jars, the heavier gas (moist air) settles more at the bottom of the jar, and has less random motion hence is more compressed and denser, than dry air that has more freedom to move randomly because of its lesser weight.

4 0
3 years ago
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