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3 years ago

13

Please read and an 3. Many Jacks use hydraullc power. A) O True B) O False

1 answer:

drek231 [11]3 years ago

5 0

Answer:

A) True. I hope this helps

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Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph

kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

$S.G=\frac{\rho_{sw}}{\rho_{w}}$

Here, density of sea water is $\rho_{sw}$ , surface gravity is S.G and density of water is $\rho_{w}$ .

Substitute all the values in the above equation as follows:

$S.G=\frac{\rho_{sw}}{\rho_{w}}$

$1.025=\frac{\rho_{sw}}{1000}$

$\rho_{sw}=1025$ kg/m³.

Step2

Difference in pressure is calculated as follows:

$\bigtriangleup p=rho_{sw}gh$

$\bigtriangleup p=1025\times9.81\times320$

$\bigtriangleup p=3217680$ pa.

Or

$\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})$

$\bigtriangleup p=3217.68$ kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

6 0

3 years ago

You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19

Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

$V = A*t$

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

$V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}$

Now, using the density we can find the mass:

$m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g$

Finally, with the Avogadros number ( $N_{A}$ ) and with the atomic mass (A) we can find the number of atoms (N):

$N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms$

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0

3 years ago

True or false. Part of the mission of the NTSB is to determine the probable cause of an accident

eimsori [14]

uniform

welcome 2 Ghana African state western region

6 0

3 years ago

Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have fou

Bingel [31]

Ucsaaaaauxx627384772938282’cc ed un e uff ridicolizzarla +golfista

4 0

3 years ago

Hareem and Shahad are on a road trip to Florida. They pull over to get gas, and as you may know it is sold by the gallon in the

Alona [7]

Answer:

53.3

Explanation:

4×14=56

56÷1.05=53.3

4 0

3 years ago