Hi! Let me help you!
a = (Vf - Vi)/t ; where distance d = [2(t)]/(4+t), t = 5secs, and Vi = 0
a = [(2t)/(4+t)]/t <---- working equation
a = {[2(5)]/9}/5 <---- cancel 5
a = 2/9 ft/s^2 <---- Answer
Answer:
The image distance is 30 cm
image height = - 5 cm
Explanation:
The formula for calculating the image distance is expressed as
1/f = 1/u + 1/v
where
f is the focal length
u is the object distance
v is the image distance
From the information given,
u = 30
f = 15
By substituting these values into the formula,
1/15 = 1/30 + 1/v
1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30
Taking the reciprocal of both sides,
v = 30
The image distance is 30 cm
magnification = image height/object height = - v/u
Given that object height = 5 cm, then
image height/5 = - 30/30 = - 1
image height = - 5 * 1
image height = - 5 cm
Answer:
a) 
b) 
c) 
Explanation:
From the question we are told that:
Given Frequencies
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
Generally the equation for Waveform Period is mathematically given by
Therefore
a)
For
b)
For
c)
For
