Answer:
a) 351351.35m/s
b) 1.044*10^{-8}kg/C
Explanation:
a) Electric force and magnetic force over the charge must have the same magnitude. From there we can compute the seep of the charge.

b) the mass-charge ratio is given by:

hope this helps!!
Answer:
1. d. a nucleus
2. b. Membrane-bound organelles
3. c. ten time smaller
4. b. animal cell
5. c. Bacteria
Explanation:
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Answer: The final mass of the tank is 2.46kg
Explanation: All shown in the attachment.
Assumptions:
i. Argon is treated as an ideal gas at the specified conditions.
ii. Isentropic relation of ideal gas applies at the given conditions.
Complete Question
The complete question is shown on the first uploaded image
Answer:
a

b

Explanation:
From the question we are told that
The amplitude is 
The period is 
The test weight is 
Generally the radial acceleration is mathematically represented as

at maximum angular acceleration

So

Now
is the angular velocity which is mathematically represented as

Therefore
![a_{max} = [\frac{2 * \pi}{T} ]^2 * A](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20%20%5B%5Cfrac%7B2%20%2A%20%20%5Cpi%7D%7BT%7D%20%5D%5E2%20%2A%20A)
substituting values
![a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20%20%5B%5Cfrac%7B2%20%2A%20%203.142%7D%7B17%7D%20%5D%5E2%20%2A%200.018)

Generally this test weight is mathematically represented as
Where k is the spring constant
Therefore

substituting values


To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

Where
B= Magnetic Field
A = Area
Angle between magnetic field lines and normal to the area
The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.
In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:
0 then our expression can be written as

From the same value of the electromotive force we have to

Replacing we have

Replacing with our values we have that


Therefore the magnitude of the induced emf in the loop is 0.0237V
On the other hand we have that the current by Ohm's Law can be defined as

For the given value of the resistance and the previously found potential we have to

