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3 years ago

The rate of doing work is called?

Physics

2 answers:

Lorico [155]3 years ago

8 0

The rate of doing work is called power

Hope I helped
jcossin22

iragen [17]3 years ago

7 0

- - The rate, according to physics, of doing work, is called "power"

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Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

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3 years ago

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Do cheese and chips good?

olga_2 [115]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A piston–cylinder device initially contains 0.22 kg of steam at 200 kPa and 300°C. Now, the steam is cooled at constant pressure

wel

Explanation:

Defining compressibility factor (Z), which is also known as the gas deviation factor, is a correction factor which describes the deviation of a real gas from ideal gas behaviour.

Z = p/(density * Rs *T)

Where Rs is specific gas constant =R/M

Deriving the volume values from steam tables,

At 200kPa and 300°C, V1 = 1.31623 m3/kg

At 200kPa and 150°C, V2 = 0.95986 m3/kg

Change in volume, DV = m(V1 - V2)

= 0.2*(1.31623 - 0.95986)

= 0.07128 m3/kg

Also from steam tables,

The critical pressure, Pc = 22.06MPa

The critical temperature, Tc = 647.1K

Remember, gas constant, R = 0.4615 kPa.m3/kg.K

Therefore, reduced temperature at initial state, Tr = T1/Tc

= (360 + 273)/647.1

= 0.886

Also, reduced pressure at initial state, Pr = P1/Pr

= 200 x 10^-3/22.06

= 0.0091

But compressibility factor at initial state, Z1 = 0.9956

Ideal volume, Vi = m*R*T/P1

= (0.2*0.4615*(300+273))/200

= 0.2644m3

Compressibility volume, Vc = Z1*Vi

= 0.9956* 0.2644

= 0.2633m3

Reduced temperature at final state, Tr = T2/Tc

= (150+273)/647.1

= 0.65

Reduced pressure at final state, Pr = P2/Pc

= 200 x 10^-3/22.06

= 0.0091

Compressibility factor at final state, Z2 = 0.9897

3 0

3 years ago