A pendulum is used in a large clock. The pendulum has a mass of 2kg. If the pendulum is moving at a speed of 2.9 m/s when it rea
2 answers:
Answer : Maximum height of the pendulum is 0.43 m
Explanation :
Given that,
Mass of the pendulum, m = 2 kg
Speed of the pendulum, v = 2.9 m/s
When it reaches the lowest point on its path, the potential energy at the highest point is equal to the kinetic energy at the lowest point.
i.e.
h is the height of the pendulum.
Hence, this is the required solution.
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Answer:
a)
b)
c) Compressing is easier
Explanation:
Given:
Expression of force:
where:
when the spring is stretched
when the spring is compressed
hence,
a)
From the work energy equivalence the work done is equal to the spring potential energy:
here the spring is stretched so,
Now,
The spring constant at this instant:
Now work done:
b)
When compressing the spring by 0.05 m
we have,
<u>The spring constant at this instant:</u>
Now work done:
c)
Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.
Answer:
26 days
Explanation:
m = 9.4×1021 kg
r= 1.5×108 m
F = 1.1×10^ 19 N
We know Fc =
==> 1.1 × = (9.4 ×
×
) ÷ 1.5 ×
==> 1.1 × =
× 6.26×
==> = 1.1 ×
÷ 6.26×
==> = 0.17571885 ×
==> v= 0.419188323 × m/sec
==> v= 419.188322834 m/s
Putting value of r and v from above in ;
T= 2πr ÷ v
==> T= 2×3.14×1.5× ÷ 0.419188323 ×
==> T = 22.472× 100000 = 2247200 sec
but
86400 sec = 1 day
==> 2247200 sec= 2247200 ÷ 86400 = 26 days
Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:
where
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:
where