A small ball is attached to one end of a rigid rod with negligible mass. The ball and the rod revolve in a horizontal circle wit
1 answer:
Answer:
mass of the ball will be 1 kg
Explanation:
We have given force F = 0.5 N
This force must be equal to the centripetal force for the ball to revolve in the circle
So
Centripetal acceleration is given as
We have to calculate the mass of the ball
We know that force is given by
So
m = 1 kg
So mass of the ball will be 1 kg
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Answer:
61.85 ohm
Explanation:
L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm
Let ω0 be the resonant frequency.
ω0 = 2357 rad/s
ω = 2 x 2357 = 4714 rad/s
XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm
Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm
Impedance, Z =
Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm
Thus, the impedance at double the resonant frequency is 61.85 ohm.
Answer:
A) 8,368 J
B) ) 0.893 J/gºC
Explanation:
A)
- The heat gained by the water can be obtained solving the following equation:
- where cw = specific heat of water = 4.184 J/gºC
- m= mass of water = 1,000 g
- ΔT = 2ºC
- Replacing these values in (1) we get:
B)
- Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.
- Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:
⇒
- which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.