Answer:
1.7333333m/s²
Explanation:
Tension of the line = the weight + force from pulling up the fish
30N = mg + ma
30 = (6)(9.8) + (6)a
10.4 = 6a
∴ a = 1.7333333m/s²
Answer: 
Explanation:
Given
Diameter of the rod 
length of rod is 
Resistivity of silicon is 
cross-section of the rod 

Resistance of rod is R


Current is given by

Answer : The final temperature is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of ice = 
= specific heat of water = 
= mass of ice = 50 g
= mass of water = 200 g
= final temperature = ?
= initial temperature of ice = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the final temperature is, 
Answer:
the work that must be done to stop the hoop is 2.662 J
Explanation:
Given;
mass of the hoop, m = 110 kg
speed of the center mass, v = 0.22 m/s
The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;
W = ΔK.E
W = ¹/₂mv²
W = ¹/₂ x 110 x 0.22²
W = 2.662 J
Therefore, the work that must be done to stop the hoop is 2.662 J