Answer:
Work done by the machine (W) = 500 × 1.5 = 750 J
Work supplied to the machine (W) = 100 × 10 = 1000 J
Here, work supplied to the machine is input work = 1000 J
P=w/t
w=15
t=3
therefore, 5 watts (b)
heat released Q = 749 joules
heat of fusion of silver L = 109 J/g
Here phase of silver is changing from liquid to solid
so temperature will remain same
all heat will be released due to its phase change
and in this case we use Q=mL
where m is the mass of silver in gram
Q= mL
749 = m * 109
m = 749/109
m = 6.87 gram
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
