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]}](https://tex.z-dn.net/?f=c%3D%20%5Cfrac%7BQ%7D%7Bm%5BT%27-T%5D%7D)
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.
Soothing? I don't really know are there any choices?
Most countries besides the USA use the metric system also known as S.I (system internationale) as a system of measurement...
Answer:
Where the electric potential is constant, the strength of the electric field is zero.
Explanation:
As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e
Eₓ = - dV / dx ----------(i)
From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.
<em>Therefore, a constant electric potential means that electric field is zero.</em>