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fiasKO [112]
2 years ago
9

Monochromatic light is incident on a grating that is 75 mm wide and ruled with 50,000 lines. the second-order maximum is seen at

32.5°. what is the wavelength of the incident light?
Physics
1 answer:
DIA [1.3K]2 years ago
4 0

Answer:

The wavelength of the incident light is \lambda = 400 nm

Explanation:

Given data

Distance between the sits

d = \frac{0.075}{50000}

d = 1.5 × 10^{-6} m

\theta = 32.5°

m = 2

We know that the wavelength of the incident light is given by

\lambda = \frac{d\sin \theta}{m}

Put all the value in above formula we get

\lambda = \frac{1.5 (\sin 32.5)}{2}×10^{-6}

\lambda = 4 × 10^{-7} m

\lambda = 400 nm

Therefore the wavelength of the incident light is \lambda = 400 nm

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Use the formula a=h/n to find the batting average a of a batter who has h hits in n times at bat. Solve the formula for h. if a
Inessa [10]
A = h / n => h = a*n

a = 0.290 hit / time
n = 300 times

=> h = 0.290 hit / time * 300 time = 87 hits

Answer: 87 hits
4 0
2 years ago
A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o
sergejj [24]

Answer:

kinetic friction or something else

Explanation:

Draw a diagram to illustrate the problem, as shown below.

m = mass of the box, kg

mg =  weight of the box, N, acting downward

Because of Newton's 3rd law, a normal force of N = mg acts upward on the box.

C = 200 N, the force applied by Clara, horizontally to the right

E = 500 N, the force applied by Eduardo, horizontally to the left.

Because E is greater than C, the box will move left.

The frictional force, F, will resist the initial motion. Its magnitude is

F = μN = μmg, where μ = kinetic coefficient of friction.

F acts horizontally to the right.

From the given table, we can conclude that

Normal force acting upward is correct

Tension by Eduardoacting left is correct.

Tension by Clara acting right is correct.

Kinetic friction acting left is Incorrect.

Gravity acting downward is correct.correct

Answer:

The direction of the kinetic friction force is incorrect

8 0
2 years ago
A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference,
Stels [109]

Answer:

<h2>0.056 W</h2>

Explanation:

Power = IV

From ohms law we know that

V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}

\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W

<em>The  resistor will dissipate 0.056 Watt</em>

7 0
2 years ago
A combined circuit has two resistors in series (27.5 ohms and 33.0 ohms) and another in parallel (22.0 ohms). If the power sourc
Dafna1 [17]
R 1,2 = 27.5 + 33.0 = 60.5 Ohms
1/ R 1,2,3 = 1/ 60.5 + 1 / 22 = 82.5 / 1331
R 1, 2, 3 = 1331 / 82.5 = 16.13 Ohms
I = U / R
I = 9 V / 16.13 Ohms = 0.557 A ≈ 0.56 A
Answer: C ) 0.56 Amps 
4 0
2 years ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
2 years ago
Read 2 more answers
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