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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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dalvyx [7]

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

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

\dfrac{d\phi}{dt} is the rate of change if magnetic flux.

\phi=BA\ cos\theta

\theta is the angle between the magnetic field and the normal to area vector.

\theta=90-60=30

\epsilon=NA\dfrac{dB}{dt}\times cos30

\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)

\epsilon=0.0598\ T

\epsilon=59.8\ mT

or

EMF = 60 mT

So, the magnitude of  emf induced in the coil is 60 mT. Hence, this is the required solution.

6 0
3 years ago
Two spherical asteroids have the same radius R. Asteroid 1 hasmass M and asteroid 2 has mas 2M. The two asteroids are releasedfr
gtnhenbr [62]

Answer:

 v_1 =\sqrt{\dfrac{16GM}{15R}}

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Explanation:

given,

mass of asteroid 1 = M

mass of asteroid 2 = 2M

radius of two asteroid = R

Distance between the asteroid = 10 R

Speed of the asteroid before collision = ?

using conservation of momentum

M u + 2M u' = M v₁ + 2 M v₂

initial speed of asteroid is equal to zero

0 = v₁ + 2 v₂

v₁ = -2 v₂

using conservation of momentum

initial potential energy is converted into potential energy and the kinetic energy of both the asteroids.

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 6v_2^2 = \dfrac{8GM}{5R}

 v_2 =\sqrt{\dfrac{4GM}{15R}}

now,

 v_1 =-2\sqrt{\dfrac{4GM}{15R}}

 v_1 =\sqrt{\dfrac{16GM}{15R}}

hence, the velocity of asteroid are

 v_1 =\sqrt{\dfrac{16GM}{15R}}

 v_2 =\sqrt{\dfrac{4GM}{15R}}

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A 1 uF parallel-plate capacitor is charged to 12 V and then disconnected from the voltage supply. The plate separation distance
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Answer:

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