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.
Answer:
6m/s
Explanation:
Given parameters:
Initial velocity = 0m/s
Acceleration = 2m/s²
Distance = 9m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we use the expression below:
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
v² = 0² + (2 x 2 x 9) = 36
v = 6m/s
Answer:
Charge on each is 2 x 10⁻¹⁰.
Explanation:
We know that Force between two point charges is given b the Coulomb's law as:
F = kq₁q₂/r^2
k = 9 x 10^9
r = 3.00 cm
= 0.03 m
q₁ = q₂
F = 4.00 x 10^-7
Rearranging the formula, we get:
F = k q²/r²
q² = Fr²/k
q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)
q² = 4 x 10⁻²⁰
q = 2 x 10⁻¹⁰
As there is force of repulsion between the charges, the charges must be both positive or both negative.
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Answer: Option (b) is the correct answer.
Explanation:
Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.
Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.
Thus, we can conclude that negative charge spread evenly on both ends.