a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
and solving for t we find
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
And by solving for t, we find
Answer:
-Relative refractive index– It is the ratio of speed of light in one medium to the speed of light in another medium
-Absolute refractive index– It is the ratio of light in vacuum to the speed of light in another medium.
Explanation:
Answer:
Explanation:
Height of building
H = 6m
Horizontal speed of first balloon
U1x = 2m/s
Second ballot is thrown straight downward at a speed of
U2y = 2m/s
Time each gallon hits the ground
Balloon 1.
Using equation of free fall
H = Uoy•t + ½gt²
Uox = 0 since the body does not have vertical component of velocity
6 = ½ × 9.8t²
6 = 4.9t²
t² = 6 / 4.9
t² = 1.224
t = √1.224
t = 1.11 seconds
For second balloon
H = Uoy•t + ½gt²
6 = 2t + ½ × 9.8t²
6 = 2t + 4.9t²
4.9t² + 2t —6 = 0
Using formula method to solve the quadratic equation
Check attachment
From the solution we see that,
t = 0.9211 and t = -1.329
We will discard the negative value of time since time can't be negative here
So the second balloon get to the ground after t ≈ 0.92 seconds
Conclusion
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
Answer:
ε= 7.86 mV, Current: Anti-clockwise
Explanation:
radius= 50 mm
dt= 0.10 s
Initial magnitude of magnetic field= B1 = 200 mT
Final magnitude of magnetic field = B2 = 300 mT
Ф= B. A= BA cosα
Ф1= B1 * A * cosα
Ф1= 
Ф1= 0.00157 Wb
Ф2= B2 * A * cosα
Ф1= 
Ф2=0.00236 Wb
dФ= Ф2 - Ф1
dФ= 0.00236 - 0.00157
dФ= 0.000786 Wb
ε= 
Ф
ε=0.001786/ 0.10
ε=0.00786 v = 7.86 mV
b)
According to lenz's law the induced emf always oppose the cause producing it.
Applied field is out of the paper so the current will flow in anti-clockwise direction to produce north pole pointing toward the paper.
