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Debora [2.8K]
1 year ago
11

In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What

is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Physics
1 answer:
Alex777 [14]1 year ago
7 0

Answer:

λ₂ = 357.3 nm

Explanation:

The expression for double-slit interference is

          d sin θ = m λ                 constructive interference

          d sin θ = (m + ½) λ        destructive interference.

The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits

         d sin θ = m λ₁

       

now ask for destructive interference for m = 4

        d sin θ = (m + ½) λ₂

we match these two expressions

         m λ₁ = (m + ½) λ₂

         λ₂ = ( m / m + ½) λλ₁  

let's calculate

         λ₂ =\frac{4}{(4.000 +0.5) \ 401}

        λ₂ = 357.3 nm

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A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
love history [14]

Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α   Formula  (1)

where:

τ : It is the moment applied to the body.  (Nxm)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :  

ωf = ω₀+ α*t  Formula (2)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t : time interval (rad)

Data  

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel  

We replace data in the formula (2):  

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

τ = F *R

Where:

F  = tangential force (N)

R  = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

8 0
1 year ago
A brick and a feather fall to the earth at their respective terminal velocities. which object experiences the greater force of a
horsena [70]
The brick, even though the brick would end up traveling faster, it most likely has a larger surface area therefore it would have more air resistance.
6 0
1 year ago
Read 2 more answers
You decide to work at a heart rate of 150 instead of 120. What area of F.I.T.T. did you change?
Rina8888 [55]

Key concepts

Heart rate

Exercising

The heart

Cardiovascular system

Health

Introduction

As Valentine's Day approaches, we're increasingly confronted with "artistic" images of the heart. Real hearts hardly resemble to two-lobed shapes adorning cards and candy boxes this time of year. And the actual shape of the human heart is important for its function of supplying blood to the entire body. You have likely noticed that your heart beats more quickly when you exercise. But have you ever taken the time to observe how long it takes to return to its normal rate after you're done exercising? In this science activity you'll get to do some exercises to explore your own heart-rate recovery time.

Background

Your heart is continuously beating to keep blood circulating throughout your body. Its rate changes depending on your activity level; it is lower while you are asleep and at rest and higher while you exercise—to supply your muscles with enough freshly oxygenated blood to keep the functioning at a high level. Because your heart is also a muscle, exercise, in turn, helps keep it healthy. The American Heart Association recommends that a person does exercise that is vigorous enough to raise their heart rate to their target heart-rate zone—50 percent to 85 percent of their maximum heart rate, which is 220 beats per minute (bpm) minus their age for adults—for at least 30 minutes on most days, or about 150 minutes a week in total. So for a 20-year-old, the maximum heart rate would be 200 bpm, with a target heart-rate zone of 100 to 170 bpm. (For those 19 or younger, target zones can vary more than they do for adults.)

i think it will help you...if it help you ...please mark brainless

8 0
2 years ago
Glider‌ ‌A‌ ‌of‌ ‌mass‌ ‌0.355‌ ‌kg‌ ‌moves‌ ‌along‌ ‌a‌ ‌frictionless‌ ‌air‌ ‌track‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.095‌ ‌m/s.‌
NemiM [27]

Answer:

vB' = 0.075[m/s]

Explanation:

We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.

Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

(m_{A}*v_{A})+(m_{B}*v_{B})=(m_{A}*v_{A'})+(m_{B}*v_{B'})

where:

mA = 0.355 [kg]

vA = 0.095 [m/s] before the collision

mB = 0.710 [kg]

vB = 0.045 [m/s] before the collision

vA' = 0.035 [m/s] after the collision

vB' [m/s] after the collison.

The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.

(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]

7 0
1 year ago
A pitcher delivers a fast ball with a velocity of 43 m/s to the south. the batter hits the ball and gives it a velocity of 51 m/
hoa [83]
Assuming north as positive direction, the initial and final velocities of the ball are:
v_i=-43 m/s (with negative sign since it is due south)
v_f=+51 m/s
the time taken is t=1.0 ms=0.001 s, so the average acceleration of the ball is given by
a= \frac{v_f-v_i}{t}= \frac{51 m/s-(-43 m/s)}{0.001 s}=9.4 \cdot 10^4 m/s^2
And the positive sign tells us the direction of the acceleration is north.
4 0
1 year ago
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