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4 years ago

In a light interference pattern, alternate dark and bright fringes are seen. Why are the bright fringes formed?

Physics

2 answers:

murzikaleks [220]4 years ago

8 0

The correct choice is

d) constructive interference

bright fringes are formed as a result of constructive interference between the two waves. Constructive interference takes place between the two waves when crest of one wave falls on crest of another wave or trough of one wave falls on trough of another wave. the new wave thus produced has higher amplitude than any of the two waves. that is how we get bright fringes.

madreJ [45]4 years ago

7 0

The light interference pattern is a series of light and dark stripes.

The bright stripes are areas where there is CONstructive interference.

The dark stripes are areas where there is DEstructive interference.

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A 1500 kg elevator, suspended by a single cable with tension 16.0 kN, is measured to be moving upward at 1.2 m/s. Air resistance

kirill115 [55]

Tension in the Cable is 0.87 m/s6^2, Elevator's speed after it has moved 10m is 1.6*10^5 N, Work done by gravity is 1.47 * 10^3 N, Elevator Kinetic Energy is 13897.5J and Elevator Speed after rising to 10m is 4.312 m/s.

Tension is a pulling force that operates in one dimension along the cables' axes in the opposite direction from the direction of the applied force. The combined weight of the elevator box and the passenger riding inside it, in the case of an elevator, provides the pulling force in the cables is called Tension.

A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. The type of motion can be vibration, translation, rotation around an axis, or any combination of these. Kinetic energy is a type of energy that an item or particle possesses as a result of motion.

We know that,

Tension in the Cable

T = m(g+a) = g+a = T/m = 16 * 103 / 1500 = 10.67 m/s2

a = 10.67 - 9.8 = 0.87 m/s6^2

Elevator's speed after it has moved 10m.

U^2 = u^2 +2as

= 1.22 +2*0.87*10

=1.6*10^5 N

Work done by gravity = mg * 10 = 14700 * 10 = 1.47 * 10^3 N

Elevator Kinetic Energy = 1/2 mv^2 = 1/2*1500*18.53 = 13897.5J

Elevator Speed after rising to 10m ,

U^2 = u^2 +2as = 1.2 +2*0.87*10 = 18.6

U =(18.6)^1/2=4.312 m/s

Learn more about Kinetic Energy here

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6 0

1 year ago

timama [110]

Answer:

Workdone = 1960 Joules.

Explanation:

Given the following data;

Mass = 5kg

Force = 49N

Height (distance) = 40m

To find the workdone;

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 49*40

Workdone = 1960 Joules.

Therefore, the amount of work done on the bowling ball to lift it is 1960 Joules.

7 0

3 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t

coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

Y axis

N- W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.32 9.8

a = - 3.14 m/s²

We calculate the distance using the kinematics equations

Vf² = Vo² + 2 a x

x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

x = ( 0 - 15²) / 2 (-3.14)

x= 35.8 m

The shortest braking distance is 35.8 m

7 0

3 years ago

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

7 0

4 years ago

Two spaceships leave Earth in opposite directions, each with a speed of 0.60c with respect to Earth. (a) What is the SPEED of sp

Studentka2010 [4]

Answer:

The relative speed of 1 relative to 2 is 0.88c

Explanation:

In relativistic mechanics the relative speed between 2 objects moving in different direction is given by

$v_{ab}=\frac{v_{a}+v_{b}}{1+\frac{v_{a}v_{b}}{c^{2}}}$

Since it is given that

$v_{a}=0.6c\\\\v_{b}=0.6c$

Applying values in the formula we get

$v_{ab}=\frac{0.6c+0.6c}{1+\frac{(0.6c)^{2}}{c^{2}}}\\\\v_{ab}=0.88c$

8 0

4 years ago