Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
Answer:

Explanation:
First of all, we need to calculate the total energy supplied to the calorimeter.
We know that:
V = 3.6 V is the voltage applied
I = 2.6 A is the current
So, the power delivered is

Then, this power is delivered for a time of
t = 350 s
Therefore, the energy supplied is

Finally, the change in temperature of an object is related to the energy supplied by

where in this problem:
E = 3276 J is the energy supplied
C is the heat capacity of the object
is the change in temperature
Solving for C, we find:

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.
A would be the correct answer. Its the only one to make sense since you are trying to solve the conflict!
<span>The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the "Radio waves"
So, option B is your answer
Hope this helps!
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