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Diano4ka-milaya [45]
3 years ago
11

What problems would we have if Pascal had failed to give the Pascal's law? Write some points.

Physics
2 answers:
Reika [66]3 years ago
4 0
Hydraulic jacks, automobile brakes and even the lift generated on airplane wings can be explained using Pascal's principle. Pascal's principle is based on the idea that fluids at rest are incompressible, allowing very large forces to be transmitted with the application of a smaller force.
Gemiola [76]3 years ago
4 0

Answer:

Pascal's law states that a change in pressure at any point in an enclosed fluid is transmitted equally throughout the fluid.

Explanation:

Applications of Pascal's law include

>The hidden standard of the water powered jack and pressure driven press.  

>Power intensification in the slowing mechanism of most engine vehicles.  

>Utilized in artesian wells, water pinnacles, and dams.  

>Scuba jumpers must comprehend this rule. At a profundity of 10 meters submerged, pressure is double the environmental weight adrift level, and increments by around 100 kPa for each expansion of 10 m depth.[5]  

>Generally Pascal's standard is applied to bound space (static flow), however because of the consistent flow process, Pascal's rule can be applied to the lift oil instrument (which can be represented as a U tube with pistons on either end).

Without Pascal's law, none of the above could be possible.

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Answer:

true

Explanation:

this is fair to other people

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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow
Oksana_A [137]

Answer:

a)\omega=1.36rad/s

b)\omega=12.99rpm

c)F=705.6N

Explanation:

a) The angular velocity is related to the centripetal acceleration by the formula a_{cp}=\omega^2r, which for our purposes we will write as:

\omega=\sqrt{\frac{a_{cp}}{r}}

Since <em>we want this acceleration to be 1.5 times that due to gravity</em>, for our values we will have:

\omega=\sqrt{\frac{1.5g}{r}}=\sqrt{\frac{(1.5)(9.8m/s^2)}{(8m)}}=1.36rad/s

b) 1 rpm (revolution per minute) is equivalent to an angle of 2\pi radians in 60 seconds:

1\ rpm=\frac{2\pi rad}{60s} =\frac{\pi}{30}rad/s

Which means <em>we can use the conversion factor</em>:

\frac{1\ rpm}{\frac{\pi}{30}rad/s}=1

So we have (multiplying by the conversion factor, which is 1, not affecting anything but transforming our units):

\omega=1.36rad/s=1.36rad/s(\frac{1\ rpm}{\frac{\pi}{30}rad/s})=12.99rpm

c) The centripetal force will be given by Newton's 2nd Law F=ma, so on the centripetal direction for our values we have:

F=ma=(48kg)(1.5)(9.8m/s^2)=705.6N

8 0
3 years ago
An astronaut takes a book onto the space shuttle. An identical book remains on earth . Which statements about the pull of earth'
worty [1.4K]
B. Earth's gravity has a stronger pull on the book when it is on earth than when it is in space.

When objects move farther away from the earth’s, the force of gravity on these objects gradually decline although their mass remains constant.
7 0
3 years ago
Will give brainliest to correct answer.
Darina [25.2K]

Answer:

1.5X10^-4C

Explanation:

Expression for the electric force between the two charges is given by -

F = (k*q1*q2) / r^2

Here, k = constant = 9 x 10^9 N*m^2 / C^2

q1 = unknown

F=25N

q2 = 1.9x10^-6 C

r = 0.32 m

Substitute the given values in the above expression -

25=9x10^9 *(q1) * 1.9 x10^ -6/ (0.32m)^2

25= 17,100* (q1)/ 0.1024

multiply both sides by 0.1024 to get rid of the denominator

2.56=17,100*(q1)

Divide both sides by 17,100 to isolate q1

q1=1.5x10^-4C

4 0
3 years ago
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