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I am Lyosha [343]
3 years ago
12

Josh did an experiment recording the changes in temperature in sand and water when exposed to a light source, and then when the

light source was removed. Josh noticed that the sand heated up faster than the water and then cools faster than the water. Josh concludes that
Physics
2 answers:
Marrrta [24]3 years ago
7 0

Before going to solve this question first we have to understand specific heat capacity of a substance .

The specific heat of a substance is defined as amount of heat required to raise the temperature of 1 gram of substance through one degree Celsius. Let us consider a substance whose mass is m.Let Q amount of heat is given to it as a result of which its temperature is raised  from T to T'.

Hence specific heat  of a substance is calculated as-

                                              c= \frac{Q}{m[T'-T]}

Here c is the specific heat capacity.

The substance whose specific heat capacity is more will take more time to be heated up to a certain temperature as compared to a substance having low specific heat which is to be heated up to the same temperature.

As per the question John is experimenting on sand and water.Between sand and water,water has the specific heat 1 cal/gram per degree centigrade which is larger as compared to sand.Hence sand will be heated faster as compared to water.The substance which is heated faster will also cools faster.

From this experiment John concludes that water has more specific heat as compared to sand.

frozen [14]3 years ago
3 0

Answer:

we can conclude that emissivity of sand will me more than the emissivity of water

Explanation:

As we know that here Source of light is the energy source from which energy is continuously incident on the Sand and Water

So here we know that rate of energy incident on sand and water will be from same source at same rate

Here this energy absorbed by the system is given as

\frac{dQ}{dt} = \sigma e A(T^4 - T_s^2)

so here the rate of energy absorbed is more sand than water

Also when source is off than rate of energy dissipated by sand is more than the water

So here we can conclude that emissivity of sand will me more than the emissivity of water

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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
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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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