The spring force = kx = mg.
k is spring constant = 200 N/m
x is the spring elongation = 0.2 m
F = 200×0.2= 40 N
Now, F = mg wil be the weight of the object
Thus, weight is 40 N.
Thus, now we can also find the mass of object, m = F/g = 40 N/9.8
Because g is acceleration due to gravity = 9.8 m/sec ^2
m = 4.08 kg
She puts each block of ice in the same 3000 mL beaker, each with 2000 mL of water at room temperature, and measures the temperature before and after adding ice. Therefore, small blocks of ice will have the same temperature.
Joanna puts two blocks of ice (one larger than the other) into separate cups and fills each with water. She compares the final water temperature of the two cups after each block of ice melts.
Put each block of ice in the same 3000 mL beaker, each at room temperature, put 2000 mL of water in it, and measure the temperature before and after adding ice. This way you keep the water at the same temperature in the beginning, then the temperature changes after you add the ice, giving you a better idea of the final temperature reading.
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Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
