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UNO [17]
3 years ago
11

A laser beam is incident on two slits with a separation of 0.180 mm, and a screen is placed 5.05 m from the slits. If the bright

interference fringes on the screen are separated by 1.62 cm, what is the wavelength of the laser light
Physics
1 answer:
gladu [14]3 years ago
6 0

Answer:

The answer is "530 nm".

Explanation:

d=0.180 \ nm=0.180 \times 10^{-3} \ m\\\\L= 5.05\ m\\\\\Delta y=  1.62\ cm= 1.62 \times 10^{-2} \ m\\\\

Using formula:

\Delta y=\frac{\lambda L}{d}\\\\\lambda =\frac{d\Delta y}{L}\\\\

  =\frac{0.180 \times 10^{-3} \times  1.62 \times 10^{-2} }{5.05}\\\\=5.30 \times 10^{-9} \times  \frac{1 \ nm}{10^{-9} \ m}\\\\=530 \ nm

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Describe the role of the plasma membrane
lara31 [8.8K]
The plasma membrane of a cell is a group of lipids and proteins that forms the boundary between a cell's contents and the outside of the cell.
5 0
2 years ago
The total mass of the train and its passengers is 750000kg. The train is traveling at a speed of 84m/s. The driver applies the b
fiasKO [112]

Answer:

|F| = 393750  N

Explanation:

Given that,

Total mass of the train, m = 750000 kg

Initial speed, u = 84 m/s

Final speed, v = 42 m/s

Time, t = 80 s

We need to find the net force acting on the train. The formula for force is given by :

F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{750000\times (42-84)}{80}\\\\F=-393750\ N

So, the magnitude of net force is 393750  N.

4 0
3 years ago
How much work in joules is required to lift a 23 kg box up from the ground to your waist that is 1.0 meters high, carry it 6 met
PSYCHO15rus [73]

Answer:

2682

Explanation:

Work done is given by :

Work = Force x distance

         =  mg x d

So, work done in lifting the box of 23 kg up to my waist of 1 m high is :

W = mg x d

   = 23 x 9.18 x 1

   = 211.14

Now work done carrying the box horizontally 6 meters across the room is

W = mg x d

   = 23 x 9.18 x 6

   = 1266.84

Work done in placing the box on the shelf that is 5.7 m above the ground is

W = mg x d

   = 23 x 9.18 x 5.7

   = 1203.49

So the total work done is = 211.14 + 1266.84 + 1203.49

                                          = 2681.47

                                          = 2682 (rounding off)

5 0
2 years ago
A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.13 s. W
vitfil [10]

Answer:

Part a)

T = 0.52 s

Part b)

f = 1.92 Hz

Part c)

speed = 3.65 m/s

Explanation:

As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s

so total time period of the particle is given as

T = 4\times 0.13 = 0.52 s

now we have

Part a)

T = time to complete one oscillation

so here it will move to and fro for one complete oscillation

so T = 0.52 s

Part b)

As we know that frequency and time period related to each other as

f = \frac{1}{T}

f = \frac{1}{0.52}

f = 1.92 Hz

Part c)

As we know that

wavelength = 1.9 m

frequency = 1.92 Hz

so wave speed is given as

speed = wavelength \times frequency

speed = 1.92 \times 1.9

speed = 3.65 m/s

4 0
3 years ago
Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
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