Elements that are what tend to be in the what state at room temperature.

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

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

3 years ago

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HELP I DONT HAVE THAT MUCH POINTS LEFT BUT 15 POINTSSS

Aliun [14]

Answer:

A

Explanation:

the answers are all not good, but I would say A is the best

4 0

3 years ago

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a 10 uF capacitpr is connected to a 12 V battery till its fully charged. it is then disconnected and to an uncharged 20 uF capac

likoan [24]

Answer:

Explanation:

The charge on 10μF capacitor = 10 x 12 x 10⁻⁶ = 120 μC

when it is connected with 20μF capacitor both acquires common potential whose value is

= 120 x 10⁻⁶ /( 10 +20) x 10⁻⁶ = 4 V.

Energy stored in 20μF capacitor =1/2 x 20 x 10⁻⁶ x 4 x 4 = 160 x 10⁻⁶ J.

6 0

3 years ago

In which situation is no work being done? A. a person carrying a box from one place to another B. a person picking up a box from

Anastaziya [24]

Picking up a box, pushing a box along the ground, and pulling a box along the ground (B, C, and D) definitely involve work being done.

If a person CARRIES a box from one place to another, AND keeps it at the same height during the entire carry, then no work is done. ( A )

6 0

4 years ago

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