A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill
1 answer:
Answer:
(i) The angular speed of the small metal object is 25.133 rad/s
(ii) The linear speed of the small metal object is 7.54 m/s.
Explanation:
Given;
radius of the circular path, r = 30 cm = 0.3 m
number of revolutions, n = 20
time of motion, t = 5 s
(i) The angular speed of the small metal object is calculated as;
(ii) The linear speed of the small metal object is calculated as;
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Answer:
RL=100K → Vo=9.90 mV
RL=10K → Vo=9.09 mV
RL=1K → Vo=5 mV
RL=100 → Vo=909.09 μV
In order to obtain 80% of the power source we have to put a resistor of 4 KOhm.
Explanation:
Here we have a power source in serie with a resistor of 1K and RL, in order to obtain the Vo voltage we have to apply the voltage divider rule, that states:
Substituing the resistor values of RL we obtained the following results:
RL=100K → Vo=9.90 mV
RL=10K → Vo=9.09 mV
RL=1K → Vo=5 mV
RL=100 → Vo=909.09 μV
In order to find the lowest value that gives us 80% of the source voltage we have to use the voltage divider rule again and make the Vo equal to 0.8 Vin:
The result of the last equation is 4000, so in order to obtain 80% of the power source we have to put a resistor of 4 KOhm.
Answer:
Explanation:
The relation between the de Broglie wavelength and the momentum of the particle is given by
where, m is the mas of the particle and v be the velocity of the particle and h be the Plank's constant.
So, the de broglie wavelength of proton is given by
.... (1)
The de broglie wavelength of electron is given by
.... (2)
Divide equation (2) by equation (1), we get
As the mass of proton is much more than the mass of electron, so the de broglie wavelength of electron is more than the de Broglie wavelength of proton.