Answer:
![v = 12.57 m/s](https://tex.z-dn.net/?f=v%20%3D%2012.57%20m%2Fs)
Explanation:
As we know that the radius of the circular motion is given as
![R = 50.00 cm](https://tex.z-dn.net/?f=R%20%3D%2050.00%20cm)
time period of the motion is given as
![T = 0.2500 s](https://tex.z-dn.net/?f=T%20%3D%200.2500%20s)
now we know that it is moving with uniform speed
so it is given as
![v = \frac{2\pi R}{T}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%20R%7D%7BT%7D)
now plug in all data
![v = \frac{2\pi(0.50)}{0.25}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%280.50%29%7D%7B0.25%7D)
![v = 12.57 m/s](https://tex.z-dn.net/?f=v%20%3D%2012.57%20m%2Fs)
The electrostatic force between two charges is inversely
proportional to the square of the distance between them.
So if you want to multiply the force by, say, ' Q ',
you need to multiply the distance by ( 1 / √Q ) .
We want to multiply the force by 16, so we need to
multiply the distance by ( 1 / √16 ) = ( 1 / 4 ) .
The distance should be changed to 1/4 of what it is now.
Answer:
0.2
Explanation:
The given parameters are;
The acceleration of the train, a = 0.2·g
The mass of the person standing on the train = m
Let μ represent the coefficient of static friction, we have;
The force acting on the person, F = m × a = m × 0.2·g
The force of friction acting between the feet and the floor,
= m·g·μ
For the person not to slide we have;
The force acting on the person = The force of friction acting between the feet and the floor
F = ![F_f](https://tex.z-dn.net/?f=F_f)
∴ m × 0.2·g = m·g·μ
From which we get;
0.2 = μ
The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.
The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
<h3>What is a frequency?</h3>
The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.
Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,
![\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma = 6.88 \times 10^{14}\ s^{-1}](https://tex.z-dn.net/?f=%5Crm%20%5Cgamma%20%3D%20%5Cdfrac%7BE%7D%7Bh%7D%20%3D%20%5Cdfrac%7B4.56x10%5E%7B-19%7D%20J%7D%7B6.626%20%5Ctimes%2010%5E%7B-34%7D%5C%20Jsec%5E%7B-1%7D%7D%5C%5C%5C%5C%5C%5C%5Cgamma%20%20%3D%206.88%20%5Ctimes%2010%5E%7B14%7D%5C%20s%5E%7B-1%7D)
Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
Learn more about Frequency:
brainly.com/question/5102661
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Answer:
-36.4 m/s
Explanation:
final velocity= initial velocity + acceleration x time
7 + (-9.8)(3)= -36.4 m/s