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Kobotan [32]
1 year ago
10

Suppose the wavelength of the light is 460 nm. how much farther is it from the dot on the screen in the center of fringe e to th

e left slit than it is from the dot to the right slit?
Physics
1 answer:
gavmur [86]1 year ago
3 0

The difference is 920nm.

<h3>What is called Wavelength?</h3>

The wavelength can be defined as the separation between wave crests. Additionally, a wide variety of objects move in waves, including light, the earth or ground, water, strings, air, and sound waves. Furthermore, we use the Greek letter lambda to denote the wave's wavelength. In addition, the wave's wavelength is determined by dividing its velocity by its frequency. Additionally, we employ meters (m), which is the unit used to indicate wavelength in meters.

Positive interference is produced by a bright fringe, and the wavelength is always multiple.

The difference at the center fringe is (460 nm)(0) = 0.

The difference for the first maximum is (460 nm)(1) = 460 nm.

The difference for the second maximum is (460 nm)(2) = 920 nm.

As a result, the difference for maxima n is (460 nm) (n)

Since C is on the second maximum in the image that was submitted, the difference is stated as (460 nm)(2) = 920 nm.

Consequently, the screen's central dot on the fringe E to the left slit is 920 nm.

To learn more about Wavelength, visit:

brainly.com/question/13533093

#SPJ4

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Asexual reproduction involves only one parent. The offspring of this type of reproduction have - F DNA identical to the DNA of t
Sedbober [7]

Answer:

DNA identical to the DNA of the parent

Explanation:

5 0
3 years ago
What is the answer?
Andrews [41]

Answer:

6.26 m/s

Explanation:

Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)

PE    =   KE

mgh = 1/2 mv^2     divide both sides of the equation by 'm'

gh    = 1/2 v^2             multiply both sides by 2

2 gh = v^2               take sqrt of both sides

v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s

3 0
2 years ago
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0
3 years ago
3. Differentiate between: (a) area and volume​
Margarita [4]

Answer:

area is 2d volume is 3d

Explanation:

Area refers to the size of two-dimensional surface. Volume refers to the size of a three-dimensional space.

8 0
3 years ago
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0
3 years ago
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