Coin-shaped compartment that contains light-absorbing molecules

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

Svetllana [295]

Answer:

a

$P_G = 14.03 \ psig$

b

$h_m = 0.148 \ m$

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is $P_{m} = 15.5 \ psig = 15.5 * 6894.76 = 106868.78 \ N/m^2$

The level of mercury is $h = 950 \ mm = 0.950 \ m$

The drop in the mercury level at the visible arm is $d = 39.0 = 0.039 \ m$

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

$P_g = P_m = g * \delta h * \rho$

Here $\rho$ is the density of mercury with value $\rho = 13.6 *10^{3} kg/m^3$

and $\delta h$ is the difference in the level of gas in arm one and two

So

$\delta h = \frac{106868.78}{ 13.6 *10^{3} * 9.8 }$

$\delta h = 0.802 \ m$

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

$h_m = 0.950 - 0.802$

=> $h_m = 0.148 \ m$

Generally from manometry principle we have that

$P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0$

Here $P_G$ is the pressure of the gas

$P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0$

$P_G = 9.6724 04 *10^{4} \ N/m^2$

converting to psig

$P_G = \frac{ 9.6724 04 *10^{4} }{6894.76}$

$P_G = 14.03 \ psig$

6 0

3 years ago

A ball of mass 45kg thrown upward with a velocity of 40m/s

AnnyKZ [126]

1) KE=1/2*m*v^2
1/2*45*40^2
KE=36,000J

2) PE=mgh
45*9.81*30
PE=13243.5J

4 0

3 years ago

The pressure exerted by a column of liquid is equal to the product of the height of the column times the gravitational constant

almond37 [142]

Answer:

The column of methanol is = 12.27 m high

Explanation:

The pressure of mercury (p) = ghd,................... equation 1

Where g = acceleration due to gravity, h= height, d = density.

d = 13.6 g/ml convert to kg/m³ = 13.6 g/ml(1000 (kg/m³)/(g/ml))

d =13600 kg/m³., h= 713mm = 713/1000 = 0.713 m., g= 9.80 m/s²

Substituting these values into equation 1,

P = 13600 × 0.713 × 9.8 = 95028.64 N/m².

If the pressure of the mercury supports the pressure exerted by the methanol

∴ pressure that supports the mercury = pressure of the methanol

p = ghd

making h the subject of formula,

h = p/gd ........................... equation 2

Where p = 95028.64 N/m², g = 9.8 m/s², d = 0.79 g/ml = (0.79 × 1000) kg/m³

d = 790 kg/m³.

Substituting these values into equation 2

h = 95028.64/(9.8×790)

h = 95028.64/7742

h = 12.27 m.

The column of methanol is = 12.27 m high

7 0

3 years ago

The acceleration due to gravity for any object, including 1 washer on the string, is always assumed to be m/s2. The mass of 3 wa

Pachacha [2.7K]

Answer:

The force will increase in proportion to the mass of the objects

Explanation:

The acceleration due to gravity is always the same. It is expressed in meters per second squared or m/s². The figure of 9.81 m/s² is an average value that was taken after calculating the acceleration under different surfaces. In fact, the acceleration differs depending on the shape of the part of the earth in relation to the earth's magnetic field and force.

Thus, if one washer was 20 kg, the acceleration being 9.81 m/s² the weight will be:

F = ma

= $20 * 9.81\\= 196.2 N$

If there are there washers, the weight will be:

F = 3 * 20 * 9.81

= 588.6 N

5 0

3 years ago

Read 2 more answers

Show all work and answer all 4 parts. Projectile motion. 20 points. Thank you.

Mashutka [201]

Answer:

i took g = 9.8m/s

A. 1.16secs

B. 2.32secs

C. 6.57m

D. 57.91m

Explanation:

A. How long does the Missile take to reach ot peak?

Time taken (t) =( U²Sin (angle) )/g

u = initial velocity = 25m/s

angle given = 30°

g = acceleration due to gravity = 9.8m/s²

t = U² x Sin (angle) / g

t = 25² x Sin(30)/9.8

t = 1.61secs

B. How long is the missile in the air in total?

T = 2t

T = 2 x 1.61 = 2.32 secs

C. what maximum Height does the missile reach?

Maximum height = U²Sin²(angle) / 2g
M.H =25² x Sin(30)² / 2 x 9.8
M.H=6.57m
Maximum height= 6.57m

D. How far does the missile travel Horizontally?

Range = U²2Sin(angle)/g
Range = 25² x 2 x Sin(30) / 9.8
Range = 57.91m

5 0

3 years ago