You are attending a county fair with your friend from your physics class. While walking around the fairgrounds, you discover a n

Response:

n (a sin θ) = m λ₀

Since n > 1, this indicates that the fringes separate further apart

Clarification:

In a diffraction experiment, the equation for constructive interference fringes is provided by

a sin θ = m λ₀

It is presumed that the air has been evacuated from the experiment, setting n = 1

When this experiment is conducted in water, the wavelength alters

λₙ = λ₀ / n

for achieving constructive interference

a sin θ = m λₙ

we replace

a sin θ = m λ / n

n (a sin θ) = m λ₀

Given that water's refractive index is n = 1.33, the distance between the fringes increases due to n > 1, causing the fringes to move apart

Answer:

The car that is the furthest from the finish line is: Car III (Choice C).

Explanation:

Here, we seek the car with the lowest overall average speed throughout the race. Thus, the one in last place inherently possesses the slowest average speed.

Since Car III is significantly behind Cars I and II, Choice A and B cannot be correct. Choice D is also not valid, as the positions of the cars are not the same. Lastly, Choice E is incorrect due to sufficient evidence demonstrating that Choice C has the lowest average speed.

Answer:

This excerpt is considered part of the resolution since it illustrates the events following the climax. It resolves the conflict, bringing the narrative to a conclusion.

Explanation: because I am intelligent, that's why.

Answer:

 = 3289.8 m/s

Explanation:

This problem can be approached using momentum definitions.

     I = ∫ F dt

We substitute and compute.

     I = ∫ (at - bt²) dt

Integrating gives us:

      I = a t² / 2 - b t³ / 3

We will evaluate between the limits I=0 for t = 0 ms and higher I=I for t = 2.74 ms:

      I = a (2.74² / 2- 0) - b (2.74³ / 3 -0)

      I = a 3.754 - b 6.857

Substituting the values for a and b, we find:

      I = 1500 3.754 - 20 6.857

      I = 5,631 - 137.14

      I = 5493.9 N s

Next, we engage the relationship between impulse and momentum:

      I = Δp = m - m v₀o

      I = m - 0

       = I / m

     = 5493.9 /1.67

     = 3289.8 m/s

Answer:

A flea can attain a maximum elevation of 51 mm.

Explanation:

Hello!

The following equations describe the height and velocity of the flea:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

In free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = flea's height at time t.

h0 = initial height.

v0 = starting velocity.

t = time interval.

a = flea's acceleration while jumping.

v = flea's velocity at that specific time.

g = gravitational acceleration.

Initially, we need to determine the time taken for the flea to attain a height of 0.0005 m. This will help us calculate the flea's velocity during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we assume the ground as the origin, thus h0 = 0. Since the flea starts stationary, v0 = 0. Therefore:

h = 1/2 · a · t²

We need to find the value of t when h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Next, we calculate the velocity achieved during that time:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

At a height of 0.50 mm, the flea's velocity stands at 1.00 m/s. This initial speed will reduce due to gravity's downward pull. When the speed reaches zero, the flea will have reached its peak height. Using the velocity equation, let's determine the time taken to reach maximum height (v = 0):

v = v0 + g · t

At peak height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, we can compute the height attained by the flea during this time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

A flea reaches a maximum height of 51 mm.