Answer:
Explanation:
In Michelson interferometer , two light waves from different directions are made to overlap so that fringes are formed on the screen due to interference . In it, two monochromatic and coherent light are made to overlap which have some path difference or phase difference. They form dark and bright fringes .
Now when a match stick is lit in the path of a wave , the fringes will disappear and an general illumination will be observed on the screen as the light from the lit match stick will not be coherent . Incoherent light can not form stable fringes.
Answer:
voltage divider, R₂ = 1000 R₁
measuring the output in the resistance R₁
Explanation:
Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V
in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.
To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.
If we use two resistors whose relationship is
R₂ / R₁ = 10³
R₂ = 1000 R₁
When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator
Solving this using the time, we know that range = horizontal velocity x time of flight
since
there are no horizontal forces acting on the ball, there are no
horizontal accelerations and the initial horizontal velocity of 36 cos
28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have
range = 36 cos 28 x 3.44 s = 109.3 m
Answer:
v = 0.84m/s, v(max)= 0.997m/s
Explanation:
Initial work done by the spring, where c is the compression = 0.28m:
Work lost to friction:
Energy:
(a) Solve for v:
(b) Solve 
for x:
if:
Your answer should be D
let me know if I got it wrong
Hope this helped!
