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2 years ago

A hydraulic system is used to lift a 20,000 N vehicle in an auto garage. If the vehicle sits on

Physics

1 answer:

ollegr [7]2 years ago

7 0

Answer:

Explanation:

Area covered by the vehicle on the piston A1 = 0.5 m2 Weight of the vehicle, F1 = 2000 kg × 9.8 m s-2 Area on which force F2 is applied, A = 0.03 m2 P1 = P2 ; F 1 A 1 F1A1 = F 2 A 2 F2A2 and F2 = F 1 A 1 F1A1 A2 ; F2 = (2000 × 9.8) 0.03 0.5 0.030.5 = 1176 NRead more on Sarthaks.com - https://www.sarthaks.com/957542/hydraulic-system-used-lift-2000-vehicle-in-an-auto-garage-if-the-vehicle-sits-on-piston-area

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Which muscle disorder does donna’s doctor diagnose? donna is suffering from leg pain in which the calf muscles contract involunt

Kay [80]

Donna's doctor must have diagnosed her with muscle cramps. With the given symptoms above, it is likely that she is experiencing muscle cramps. Muscle cramps most likely happen when the individual experiencing it is lack of fluid intake, causing the muscles to tighten causing pain and for the muscle to contract involuntarily.

7 0

2 years ago

Read 2 more answers

An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t

Fofino [41]

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

$P(z)=P_{0}+\rho _wgh$

Now the gauge pressure is given by

$P(z)-P_{0}=\rho _wgh$

Applying values we get

$P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa$

8 0

3 years ago

Question 2

Yakvenalex [24]

Answer:

Explanation:

6 0

2 years ago

You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.

valentinak56 [21]

Answer:

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$

radial distance from the linear charge, $r=0.09\ m$

linear charge density, $\lambda=3\times 10^{-9}\ C.m^{-1}$

We know that the electric field by the linear charge is given as:

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

$E=150\ N.C^{-1}$

Now the force on the given charge can be given as:

$F=E.q$

$F=150\times 4\times 10^{-9}$

$F=6\times 10^{-7}\ N$

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3 years ago

Select all the correct answers.

myrzilka [38]

Anything that's dropped through air is somewhat affected by air resistance. But, out of that list, the leaf and the balloon are the items that will be affected by air resistance enough so that you can plainly see it.

If you spend some time thinking about it, you can kind of understand why airplane wings and boat propellers are shaped more like leafs and balloons than like bricks and rocks.

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3 years ago