Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.
Explanation: To find the answer, we need to know more about the simple harmonic motion.
<h3>
What is simple harmonic motion?</h3>
- A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
- We have the equation of time period of a SHM as,

- Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>

- We have to find the value of m,


Thus, we can conclude that, the mass m will be 0.625kg.
Learn more about simple harmonic motion here:
brainly.com/question/28045110
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I think it is A........................................
Sorry I had the answer but it wont let me type numbers :(.:
Answer:
the mass of water is 0.3 Kg
Explanation:
since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:
Q water + Q copper = Q surroundings =0 (insulated)
Q water = - Q copper
since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature
and denoting w as water and co as copper :
m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)
m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]
We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C
if we assume that both specific heats do not change during the process (or the change is insignificant)
m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]
m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))
m w= 0.3 kg