Answer:
3.28 cm
Explanation:
To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:
r = 
Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.
The mass and charge of a proton are:
m = 1.67 * 10^-27 kg
q = 1.6 * 10^-19 C
So, we get that the radius r will be:
r = 
= 0.0328 m, or 3.28 cm.
Answer:
Keq = 2k₃
Explanation:
We can solve this exercise using Newton's second one
F = m a
Where F is the eleatic force of the spring F = - k x
Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement
F + F = m a
k₃ x + k₃ x = m a
a = 2k₃ x / m
To find the effective force constant, suppose we change this spring to what creates the cuddly displacement
Keq = 2k₃
A. because as the merry-go-round spins the child accelerates towards the center of the merry-go-round at a uniform rate.
Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is 
Explanation:
From the question we are told that
The focal length is 
This negative sign shows the the microscope is diverging light
The angular magnification is 
The distance between the objective and the eyepieces lenses is 
Generally the magnification is mathematically represented as
![m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]](https://tex.z-dn.net/?f=m%20%20%3D%20%20%5B%5Cfrac%7BZ%20-%20f_e%20%7D%7Bf_e%7D%5D%20%5B%5Cfrac%7B0.25%7D%7Bf_0%7D%20%5D)
Where 
is the eyepiece focal length of the microscope
Now making the subject of the formula
![f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }](https://tex.z-dn.net/?f=f_e%20%20%3D%20%5Cfrac%7BZ%7D%7B1%20-%20%5B%5Cfrac%7BM%20%20%2A%20%20f_o%20%7D%7B0.25%7D%5D%20%7D)
substituting values
![f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }](https://tex.z-dn.net/?f=f_e%20%20%3D%20%5Cfrac%7B%200.19%20%7D%7B1%20-%20%5B%5Cfrac%7B300%20%20%2A%20%20-0.0055%20%7D%7B0.25%7D%5D%20%7D)
Answer:
-3044.848m/s^2
Explanation:
Initial velocity = 15.7 m/s (u)
Final velocity = -14.444 m/s (v) velocity after striking moving opposite direction.
Time = 0.0099 s
Average acceleration :
change in velocity/ total time
= V - U/t
= -14.444 m/s - 15.7 m/s/0.0099 s
= -30.144/0.0099 s
= -3044.848m/s^2
