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3 years ago

580 nm light shines on a double slit

Physics

1 answer:

lesantik [10]3 years ago

6 0

Answer:

Angular position of the first minimum is 0.27 degree

Explanation:

As we know by the formula of single slit diffraction pattern

$a sin\theta = m\lambda$

here we have

a = d = 0.000125 m

for m = 1

$\lambda = 580 nm$

now we have

$0.000125 sin\theta = 580 \times 10^{-9}$

$sin\theta = 4.64 \times 10^{-3}$

$\theta = 0.27 degree$

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Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh

Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0

3 years ago

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N,

maw [93]

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant k is

43.8 N = k (0.155 m) ==> k = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

5 0

3 years ago

3N 3 N What is the net force of the box?

zepelin [54]

6N I think I’m pretty sure

3 0

3 years ago

A 60 KG skier’s racing down a course at 20 M/S what is the skiers momentum

Xelga [282]

Answer:1200kgm/s

Explanation:

Mass=60kg velocity=20m/s

Momentum=mass x velocity

Momentum=60 x 20

Momentum=1200kgm/s

6 0

3 years ago

A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential

BARSIC [14]

Answer:

$1.88\cdot 10^{-5} A$

Explanation:

The capacitance of a parallel plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$ (1)

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

$Q=CV$ (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

$J=\frac{Q}{t}$ (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

$J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}$

where we have

$A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2$

$d = 0.58 mm = 5.8\cdot 10^{-4} m$

$\frac{V}{t}=500,000 V/s$

Substituting into the equation, we find

$J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A$

6 0

3 years ago