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ludmilkaskok [199]

3 years ago

6

Light illuminates two closely spaced thin slits and produces an interference pattern on a screen behind. How will the distance b

etween the fringes of the pattern differ for red light and blue light? 1. The same spacing for both 2. Closer for red light 3. Farther apart for red light

1 answer:

umka21 [38]3 years ago

3 0

Answer:

3. Farther apart for red light

Explanation:

The distance $y$ between fringes is

$y = \dfrac{\lambda L }{d},$

where $\lambda$ is the wavelength of light, $L$ is the distance to the screen, and $d$ is the slit separation.

Now, the wavelength $\lambda_r$ of the red light is greater than for the wavelength $\lambda_b$ for blue light:

$\lambda_r> \lambda_b$ ,

which means

$\dfrac{\lambda_r L }{d}>\dfrac{\lambda_b L }{d},$

In other words, the distance between the fringes is greater for red light, which from the options is choice 3.

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The half life of uranium-235 is 4. 5 billion years. If 0. 5 half-lives have elapsed, how many years have gone by?

igor_vitrenko [27]

Answer:

2.25 billion years

Explanation:

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5 0

2 years ago

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ExtremeBDS [4]

Answer:

When the resistances are connected in the parallel the equivalent resistance will be always less than the value of the lowest resistance in the parallel circuit.

The equivalent resistance in the parallel circuit can be calculated using the following formula.

1/R=1/R1+1/R2+1/R3

1/R= 1/10+1/20+1/30

1/R=(6+3+2)/60

1/R=11/60

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3 0

3 years ago

A disk with a radius of R is oriented with its normal unit vector at an angle Q with respect to a uniform electric field. Which

beks73 [17]

As we know that electric flux is given by

$\phi = E.A$

$\phi = EAcos\theta$

so in order to increase the flux we have two options

1. By increasing the area of the disc

2. by changing the orientation of disc so the area of the disc is parallel to the electric field

so correct answer will be

<em> A. increasing the area of the disk</em>

7 0

3 years ago

Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially

statuscvo [17]

Answer:

Final speed is 900.06 m/s at $0.2215^{\circ}$

Solution:

As per the question:

Mass of the first asteroid, m = $15\times 10^{3}\kg$

Mass of the second asteroid, m' = $20\times 10^{3}\kg$

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, $\theta = 20^{\circ}$

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

$\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}$

Thus applying vector addition and momentum conservation, the final velocity is given by:

$(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}$ (1)

Now,

$(m +m')v_{final} = (35\times 10^{3})v_{final}$

$(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}$

$(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}$

$2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}$

Now, substituting the suitable values in eqn (1), we get:

$v_{final} = 900.06\ m/s$

Now, the direction for the two vectors is given by:

$\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}$

$\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}$

5 0

3 years ago

While taking the stairs it takes you 10 seconds to reach the top. The next time you take the same stairs, it takes you 5 seconds

Ksju [112]

Answer:

P₂ = 2 P₁

we conclude that in the second time the power used is double that in the first rise

Explanation:

In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.

Now we can analyze the required power,

P = W / t

From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,

therefore the first time the power is

P₁ = E / 10

P₁ = 0.1 E

for the second time the power is

P₂ = E / 5

P₂ = 0.2 E

we see that the power in the second case is

P₂ = 2 P₁

Therefore, we conclude that in the second time the power used is double that in the first rise.

6 0

3 years ago