Answer:
This question is incomplete
Explanation:
This question is incomplete because of the absence of the time taken to complete one full cycle.
Frequency (<em>f</em>) will be calculated first as
<em>f </em>= <em>N </em>÷<em> t</em>
where <em>N </em>is the number of cycles and <em>t </em>is the time taken to complete one full cycle. The unit for frequency is Hertz (Hz).
To calculate the period, <em>T, </em>the formula below will be used
<em>T </em>= 1 ÷ <em>f</em>
The unit for period is secs
I'm not sure about the distance to the nearest star, but it's probably about 4 light-years (L-y).
1 L-y = 1.86 * 10E5 mi/sec * 3600 sec/hr * 24 hr/day * 365 day/yr
1 L-y = 5.9 *10E12 mi and 4 L-y = 2.3 *10E13 mi distance to star
2.3 * 10E13 mi / 900 mi/hr = 2.6 * 10E10 hr hours to star
2.6 * 10E10 hr / (24 hr/day) = 1.1 * 10E9 day days to star
1.1 * 10E9 day / 365 day/yr = 3 * 10E6 yr = 3 million years to star
Answer:
Both experienced the same magnitude impulse
Explanation:
This is because, the impulse force is internal to the system of both the tennis ball and the bowling ball. It is an action-reaction pair. So, the force exerted on the tennis ball by the bowling ball equals in magnitude to the force exerted by the tennis ball on the bowling ball although, they are in opposite directions. This, both experienced the same magnitude impulse.
Answer:
2.068 x 10^6 m / s
Explanation:
radius, r = 5.92 x 10^-11 m
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.
centripetal force = ![\frac{mv^{2}}{r}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E%7B2%7D%7D%7Br%7D)
Electrostatic force = ![\frac{kq^{2}}{r^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bkq%5E%7B2%7D%7D%7Br%5E%7B2%7D%7D)
where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2
So, balancing both the forces we get
![\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}](https://tex.z-dn.net/?f=%5Cfrac%7Bkq%5E%7B2%7D%7D%7Br%5E%7B2%7D%7D%3D%5Cfrac%7Bmv%5E%7B2%7D%7D%7Br%7D)
![v=\sqrt{\frac{kq^{2}}{mr}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7Bkq%5E%7B2%7D%7D%7Bmr%7D%7D)
![v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B9.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%205.92%5Ctimes10%5E%7B-11%7D%7D%7D)
v = 2.068 x 10^6 m / s
Thus, the speed of the electron is give by 2.068 x 10^6 m / s.
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