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Vladimir [108]
3 years ago
13

The pressure in a compressed air storage tank is 1200 kPa. What is the tank’s pressure in (a) kN and m units; (b) kg, m, and s u

nits; and (c) kg, km, and s units?
Physics
1 answer:
elixir [45]3 years ago
8 0

Explanation:

Given data

Pressure P=1200 kPa

To find

Pressure in

(a) kN/m²

(b) kg/m.s²

(c) kg/km.s²

Solution

For Part (a)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )\\P=1200kN/m^{2}

For Part (b)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )\\P=1,200,000kg/m.s^{2}

For Part (c)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )(\frac{100m}{1km} )\\P=1,200,000,000kg/km.s^{2}

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A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
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Thus applying values we get

W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J

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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
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Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
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