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Monochromatic coherent light shines through a pair of slits. If the distance between these slits is decreased, which of the following statements are true of the resulting interference pattern?A) The distance between the maxima stays the same.B) The distance between the maxima decreases.
C) The distance between the minima stays the same.
D) The distance between the minima increases.
E) The distance between the maxima increases.

Answers

Answer 1

Answer:

Answer:

The correct statements are D and E.

Explanation:

The fringe width is given by the following formula as :

$\beta =(\lambda D)/(d)$

Here,

$\lambda$ is wavelength of light

D is distance between slit and the screen

d is slit width.

If the between these slits is decreased, the fringe width increases. As a result, the distance between the minima increases and also the distance between the maxima increases.

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A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V. Use ϵ0 = 8.85×10⁻¹² C²/N⋅m². The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
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A boat can travel in still water at 56 m/s. If the boats sails directly across a river that flows at 126 m/s. What is the boats speed relative to the ground

Answers

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

V² = 56² + 126²

V² = 19012

V = 137.9 m/s

A bridge is made with segments of concrete 91 m long (at the original temperature). If the linear expansion coefficient for concrete is 1.2 × 10−5 ( ◦C)−1 , how much spacing is needed to allow for expansion for an increase in temperature of 56◦F? Answer in units of cm.

Answers

Answer:

change in length is 3.397 cm

Explanation:

Given data

long = 91 m = 9100 cm

coefficient for concrete (a) = 1.2 × 10−5 ( ◦C)−1

temperature = 56 F = (56× 5/9) ◦C

to find out

how much spacing is needed to allow

solution

we know allow space is given by this formula

change in length = coefficient for concrete × given length × temperature .............1

put all value in equation 1

change in length = 1.2 × 10−5 × 9100 × (56× 5/9)

change in length = 3.397 cm

so change in length is 3.397 cm

Determine the tension in the string that connects M2 and M3.

Answers

thereforemass m1=4.8kg and the tension

in the horizontalspring T2=10N.

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To determine the tension in the string that connects M2 and M3, we can follow these steps:

Step 1: Identify the necessary variables. Given data (for example) could be:
- Mass of M2, which is 5 kg
- Mass of M3, which is 10 kg
- The acceleration due to gravity, which is approximately 9.8 m/s²
- The angle at which the string pulls on M2, which is 30 degrees
- Assume the system is in equilibrium, meaning there is no net acceleration, so the acceleration is 0 m/s²

Step 2: Calculate the weight of M3, which is its mass times the acceleration due to gravity. This is because weight is the force exerted by gravity on an object, which equals the object's mass times the acceleration due to gravity.

For M3, this calculation would be M3 * g = 10 kg * 9.8 m/s² = 98 N (Newtons).

Step 3: Determine the force exerted by M2 that acts along the line of the string. This won't be the full weight of M2, because the string pulls at an angle. This component of the force can be calculated using the sine of the angle, because sine gives us the ratio of the side opposite the angle (here, the force along the string) to the hypotenuse (here, the full weight of M2) in a right triangle.

The horizontal component of the force of M2 is then M2 * g * sin(30deg) = 5 kg * 9.8 m/s² * sin(30deg) = 24.5 N.

Step 4: The tension in the string is the force M3 exerts on it, which is its weight, minus the component of M2's weight that acts along the string. This is because M2 and M3 are pulling in opposite directions, so they subtract from each other.

The tension in the string is then the weight of M3, 98 N, minus the horizontal (along the string) component of M2's weight, 24.5 N.

So, the tension in the string is 98 N - 24.5 N = 73.5 N.

This is the force that the string needs to exert in order to keep M2 and M3 connected and in equilibrium.

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Which of the following organisms has an adaptation that will allow it to survive in tundra biome? *A.)Plants with roots that are short and grows sideways with hairy stems and small leaves.
B.)Plants that have broad leaves to capture sunlight and long roots to penetrate the soil.
C.)Animals with thin fur that allows them to get rid of heat efficiently.
D.)Animals with long tongues for capturing prey and sticky pads for climbing trees.

Answers

Answer:

the awnser is A becuse the hair help.

High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of the track depends on the speed of the particle and its lifetime. A particle moving at 0.993c leaves a track 1.15 mm long. What is the proper lifetime of the particle

Answers

Answer:

Lifetime = 4.928 x 10^-32 s

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s

Final answer:

To find the proper lifetime of the particle, we can use the time dilation equation and the Lorentz factor. Plugging in the given values, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Explanation:

To find the proper lifetime of the particle, we can use the time dilation equation, which states that the proper time (time experienced in the frame of reference of the particle) is equal to the time observed in the laboratory frame of reference divided by the Lorentz factor. The Lorentz factor can be calculated using the equation γ = 1/√(1 - (v/c)^2), where v is the velocity of the particle and c is the speed of light. Given that the particle is moving at 0.993c, the Lorentz factor is approximately 22.82.

Next, we can use the equation Δx = βγcτ, where Δx is the length of the track, β is the velocity of the particle in units of the speed of light (v/c), γ is the Lorentz factor, c is the speed of light, and τ is the proper lifetime of the particle. Plugging in the given values, we have 1.15 mm = 0.993 * 22.82 * c * τ. Solving for τ, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

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Three point mass particles are located in a plane: a. 3.77 kg located at the origin
b. 6.7106 kg at [(5.72 cm),(11.44 cm)]
c. 2.46181 kg at [(16.7024 cm),(0 cm)].

How far is the center of mass of the three particles from the origin? Answer in units of cm