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erastovalidia [21]
2 years ago
10

What is unit of pressure?why is it called a derived unit ? give reasons​

Physics
1 answer:
olga55 [171]2 years ago
7 0

pascal is the unit of pressure

The units of pressure is called derived units because it is simply derived from base unit which is distance and a derived unit which is force, which is derived from acceleration, a derived unit as well, and mass, a base unit. As we all know, work is defined as the force x distance. Thus making work a derived unit.Jun 23, 2020

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masha68 [24]

To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as

T = 2\pi \sqrt{\frac{l}{g}}

Where

l = Length

g = Gravity

Re-arrange to find L,

L = g (\frac{T}{2\pi})^2

Our values are given as

g = 9.81m/s

T = 10.1s

Replacing,

L = g (\frac{T}{2\pi})^2

L = (9.81) (\frac{10.1}{2\pi})^2

L = 25.348m

Therefore the height would be 25.348m

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What is Br charge after it gains an electron?
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a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T
wlad13 [49]

Explanation:

Given:F=m\ddot{x}=Fe^{-\frac{t}{T}}

Solving for \ddot{x}:

\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}

where:

T=\sqrt{\frac{m}{F}}

Integrating to get \dot{x} with initial conditions \dot{x}(0)=0:

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}

Integrating to get x with initial conditions x(0) = 0:

x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}

When t=T:

x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}

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4 0
3 years ago
An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

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for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

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solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

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