The electric force is three times as much
Explanation:
The magnitude of the electric force between two charged object is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
In this problem, let's call F the initial force between the two charges:
Later, the charge on one of the particles is tripled, so that
![q_1' = 3 q_1](https://tex.z-dn.net/?f=q_1%27%20%3D%203%20q_1)
Therefore, we can calculate the new electric force between the particles:
![F'=k\frac{q_1' q_2}{r^2}=k\frac{(3q_1)q_2}{r^2}=3(k\frac{q_1 q_2}{r^2})=3F](https://tex.z-dn.net/?f=F%27%3Dk%5Cfrac%7Bq_1%27%20q_2%7D%7Br%5E2%7D%3Dk%5Cfrac%7B%283q_1%29q_2%7D%7Br%5E2%7D%3D3%28k%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D%29%3D3F)
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In physics, power is defined as energy per unit time. You will also hear it described as work per unit time. The standard unit of measure for power is the watt, where a watt is defined as joules (energy) per second (time). This is expressed as a fraction as J/s. If you wanted to increase the power in any operation, you can either increase the energy (more joules) or reduce the time (fewer seconds).
Hello!
We can use Faraday's Law of Electromagnetic Induction to solve.
![\epsilon = -N \frac{d\Phi_B}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-N%20%5Cfrac%7Bd%5CPhi_B%7D%7Bdt%7D)
ε = Induced emf (4.08 V)
N = Number of loops (?)
= Magnetic Flux (Wb)
t = time (s)
**Note: The negative sign can be disregarded for this situation. The sign simply shows how the induced emf OPPOSES the current.
Now, we know that
is analogous to the change in magnetic flux over change in time, or
, so:
![\epsilon = N \frac{\Delta \Phi_B}{\Delta t}\\\\\epsilon = N \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20N%20%5Cfrac%7B%5CDelta%20%5CPhi_B%7D%7B%5CDelta%20t%7D%5C%5C%5C%5C%5Cepsilon%20%3D%20N%20%5Cfrac%7B%5CPhi_%7BBf%7D%20-%20%5CPhi_%7BBi%7D%7D%7B%5CDelta%20t%7D)
Rearrange the equation to solve for 'N'.
![N = \frac{\epsilon}{ \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}}](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B%5Cepsilon%7D%7B%20%5Cfrac%7B%5CPhi_%7BBf%7D%20-%20%5CPhi_%7BBi%7D%7D%7B%5CDelta%20t%7D%7D)
Plug in the given values to solve.
![N = \frac{4.08}{ \frac{9.44*10^{-5} - 2.57*10^{-5}}{0.0154}} = 914.585 = \boxed{915 \text{ coils}}](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B4.08%7D%7B%20%5Cfrac%7B9.44%2A10%5E%7B-5%7D%20-%202.57%2A10%5E%7B-5%7D%7D%7B0.0154%7D%7D%20%20%3D%20914.585%20%3D%20%5Cboxed%7B915%20%5Ctext%7B%20coils%7D%7D)
**Rounding up because we cannot have a part of a loop.
The velocity of the wave is defined by the product of wavelength and frequency. The velocity of a wave with a length of 10 m and frequency of 5hz will be 50 m/sec.
<h3>What is the velocity of the wave?</h3>
The velocity of the wave is defined by the product of wavelength and frequency. It is obtained from the formula of the wavelength which is given by;
![\rm \lambda=\frac{v}{f} \\\\ \rm v = \lambda \times f \\\\ \rm v = 10 \times 5 \\\\ \rm v = 50\ m/sec](https://tex.z-dn.net/?f=%5Crm%20%5Clambda%3D%5Cfrac%7Bv%7D%7Bf%7D%20%5C%5C%5C%5C%20%5Crm%20v%20%3D%20%20%5Clambda%20%5Ctimes%20f%20%5C%5C%5C%5C%20%5Crm%20v%20%3D%20%2010%20%20%5Ctimes%205%20%5C%5C%5C%5C%20%5Crm%20v%20%3D%2050%5C%20m%2Fsec)
Hence the velocity of a wave with a length of 10 m and frequency of 5hz will be 50 m/sec.
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Answer:
D. Alpha
This is because it's unlikely for it to reach living cells inside the body