Use this formula to find your answer...
Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms
frequency (f)=1/( Time period).
Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.
Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.
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:
I didn't know these questions sorry
Answer:
a)3.5s
b)28.57m/S
c)34.33m/S
d)44.66m/S
Explanation:
Hello!
we will solve this exercise numeral by numeral
a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

where
Vo = Initial speed
=0
T = time
g=gravity=9.81m/s^2
y = height=60m
solving for time

T=3.5s
b)The horizontal speed remains constant since there is no horizontal acceleration.
with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

c)
to find the final vertical velocity we use the equations for motion with constant velocity as follows
Vf=Vo+g.t
Vf=0+(9.81 )(3.5)=34.335m/S
d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

Answer:
n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
Explanation:
Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.
In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
= (v₂-v₁)/Δt
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
= v2/R
In the general case, both the module and the address change
a = Ra ( a_{t}^2 + a_{c}^2)