Answer:
C) upward
Explanation:
The problem can be solved by using the right-hand rule.
First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).
Now we can apply the right hand rule to the charged particle:
- index finger: velocity of the particle, to the right
- middle finger: direction of the magnetic field, out of the page
- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward
Therefore, the direction of the magnetic force is upward.
Answer:
4. Force = 178.6 Newton.
5. Acceleration = 2.28 m/s².
6. Force = 178.6 Newton.
Explanation:
4. Given the following data;
Acceleration = 3.8 m/s²
Mass = 47kg
Force = mass * acceleration
Force = 47 * 3.8
<em>Force = 178.6 Newton. </em>
5. Given the following data;
Force = 785N
Mass = 345kg
Acceleration = force/mass
Acceleration = 785/345
<em>Acceleration = 2.28 m/s²</em>
6. Given the following data;
Acceleration = 6m/s²
Force = 32N
Mass =force/acceleration
Mass = 32/6
<em>Mass = 5.33 kilograms </em>
The acceleration experienced by the particle is given by
![a=113000 g=113000 \cdot 9.81 m/s^2](https://tex.z-dn.net/?f=a%3D113000%20g%3D113000%20%5Ccdot%209.81%20m%2Fs%5E2)
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed
![\omega](https://tex.z-dn.net/?f=%5Comega)
of the particle and its distance r from the axis by the relationship
![a= \omega ^2 r](https://tex.z-dn.net/?f=a%3D%20%5Comega%20%5E2%20r%20)
In our problem,
![r=6 cm=0.06 m](https://tex.z-dn.net/?f=r%3D6%20cm%3D0.06%20m)
, so we can solve for
![\omega](https://tex.z-dn.net/?f=%5Comega)
:
![\omega = \sqrt{ \frac{a}{r} } = \sqrt{ \frac{113000 \cdot 9.81 m/s^2}{0.06 m} }=4298 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%20%5Cfrac%7Ba%7D%7Br%7D%20%7D%20%3D%20%5Csqrt%7B%20%5Cfrac%7B113000%20%5Ccdot%209.81%20m%2Fs%5E2%7D%7B0.06%20m%7D%20%7D%3D4298%20rad%2Fs%20)
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to
![( \frac{1}{2 \pi} )](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B1%7D%7B2%20%5Cpi%7D%20%29)
revolutions, while
![1 s = \frac{1}{60} min](https://tex.z-dn.net/?f=1%20s%20%3D%20%20%5Cfrac%7B1%7D%7B60%7D%20min)
. So we the conversion is
Answer:
![E=8*10^5\frac{V}{m}](https://tex.z-dn.net/?f=E%3D8%2A10%5E5%5Cfrac%7BV%7D%7Bm%7D)
Explanation:
The magnitude of the electric field between two parallel conducting plates is defined as:
![E=\frac{\Delta V}{d}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B%5CDelta%20V%7D%7Bd%7D)
Here
is the potential difference between the plates and d its separation.
The electric potential energy is defined as the product between the particle's charge and the potential difference:
![U=q\Delta V](https://tex.z-dn.net/?f=U%3Dq%5CDelta%20V)
Solving for
and replacing in the electric field formula:
![\Delta V=\frac{U}{q}\\E=\frac{U}{qd}](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cfrac%7BU%7D%7Bq%7D%5C%5CE%3D%5Cfrac%7BU%7D%7Bqd%7D)
In this case we have a double charged ion, so
:
![E=\frac{32*10^3eV}{(2e)(2*10^{-2}m)}\\E=8*10^5\frac{V}{m}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B32%2A10%5E3eV%7D%7B%282e%29%282%2A10%5E%7B-2%7Dm%29%7D%5C%5CE%3D8%2A10%5E5%5Cfrac%7BV%7D%7Bm%7D)
Wavelength = (speed) / (frequency)
The speeds of the two possible signals are equal, just like
all other forms of electromagnetic radiation.
Wavelength of 895 MHz = (3 x 10⁸ m/s) / (8.95 x 10⁸/s) = 0.335 m
Wavelength of 2560 MHz = (3 x 10⁸ m/s) / (2.56 x 10⁹/s) = 0.117 m