In Young experiment, a light with a wavelength of 410 nm falls on two slitsseparated by 0.03 mm, find the angle between the central bright band and
the first order bright band
A= 0.78
B= 0.11
C= 0.032
D= 0.62

Answers

Answer 1
Answer:

The angle between the central bright band and the first order bright band is 0.78⁰, hence option A is correct.

According to the question:

A light with a wavelength of 410 nm falls on two slits separated by 0.03 mm,

So, the angle between the central bright band and the first order bright band is,

The angle will be, θ = λ/d radian

λ = 410 nm (410 × 10 ⁻⁹ m)

d = separation of slits (0.03 mm) or 0.03 × 10 ⁻³ m

θ = \rm (410* 10^-^9)/(0.03*10^-^3 ) radian

= 0.01367 radian

= 0.01367 × (180/π)⁰

= 0.78⁰

Thus, the angle between the central bright band and the first order bright band is 0.78⁰, hence option A is correct.

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Answer 2
Answer:

Answer:

B

Explanation:


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At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?

Answers

This problem can be solved using the following relation: 1 kilometer = 6.6846e-9 AU. Since we are already given the number of kilometers, we simply have to multiply it to its equivalent in AU to solve for the equivalent distance. This is done below:

46 000 000 * 6.6846e-9 = 0.3 AU

ListenA projectile with mass m is fired with initial horizontal velocity vx from height h above level ground. Which change would have resulted in a greater time of flight for the projectile? [Neglect friction.]decreasing the mass to m/2decreasing the height to h/2increasing the initial horizontal velocity to 2vxincreasing the height to 2h

Answers

The only way to increase the time of flight is to increase the height traveled by the projectile to 2h.

The given parameters;

  • mass of the projectile, = m
  • initial horizontal velocity, = v_x

The time of flight of the projectile is calculated as follows;

h = v_yt + (1)/(2) gt^2

where;

v_y is the initial vertical velocity of the projectile = 0

h = (1)/(2) gt^2\n\ngt^2 = 2h\n\nt = \sqrt{(2h)/(g) } \n\n

  • If friction due to air resistance is neglected, the time of flight of the projectile is not affected by its mass.
  • The initial horizontal velocity is not considered in the vertical direction.

Thus, the only way to increase the time of flight is to increase the height traveled by the projectile to 2h.

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Answer:

d)

Explanation:

a) decreasing the mass to m/2. FALSE. Mass doesn't influence the equations of accelerated motion.

b) decreasing the height to h/2. FALSE. From a lower initial height the projectile will reach the ground faster.

c) increasing the initial horizontal velocity to 2vx. FALSE. The horizontal and vertical components are independent, and time of flight is dependent only on vertical conditions in this case.

d) increasing the height to 2h. TRUE. From a higher initial height the projectile will reach the ground later.

A copper block is removed from a 310C oven and dropped into 1.10kg of water at 23.0C. The water quickly reaches 33.0C and then remains at that temperature.What is the mass of the copper block? The specific heats of copper and water are 385 J/(kg?K) and 4190 J/(kg?K) respectively. Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

0.432kg or 432g

Explanation:

Copper block is removed from 310C oven.

Mass of water = 1.10kg

Initial temperature of water (θ1) = 23.0C

Final temperature of water (θ2 ) = 33.0C

Specific heat of copper= 385

Specific heat of water = 4190

Let M (copper) be the mass of copper

The amount of heat lost will be absorbed by water

Q (net) = Q(water) + Q(copper) = 0 …………………(1)

Q = MCθ

For water,

Q(water) = M(water) * C(water) * (θ2-θ1)...........(2)

= 1.1 * 4190 * (33.0 -23.0)

= 1.1 *4190 * 10

= 46090J

For copper,

Q(copper) = M(copper) * C(copper) * (33.0 - 310)...........(3)

from equation 1,

Q(copper) = -Q(water)

M(copper) * C(copper) * (33.0 - 310) = - 46090

M(copper) * 385 * (-277) = -46090

M(copper) * 106645 = -46090

M(copper) = -46090 / -106645

M(copper) = 0.432kg

M(copper) = 432g

The mass of copper =432g

Final answer:

The mass of the copper block is 3.69kg.

Explanation:

To find the mass of the copper block, we can use the equation q1 = q2. The heat gained by the water is given by q1 = mcΔT, where m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature. The heat lost by the copper block is given by q2 = mcΔT, where m is the mass of the copper block, c is the specific heat of copper, and ΔT is the change in temperature. Since the water reaches a final temperature of 33.0°C, the change in temperature is ΔT = 33.0°C - 23.0°C = 10.0°C.

Using the equation q1 = q2, we have mcΔT = mcΔT. Rearranging the equation to solve for the mass of the copper block, we get m = (m1c1ΔT2) / (c2ΔT1), where m1 is the mass of the water, c1 is the specific heat of water, ΔT2 is the change in temperature of the copper block, c2 is the specific heat of copper, and ΔT1 is the change in temperature of the water. Substituting the given values, we have m = (1.10kg * 4190 J/(kg?K) * 10.0°C) / (385 J/(kg?K) * 10.0°C), which simplifies to m = 3.69kg.

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What would be the result of an alpha particle coming into a magnetic field?A) The alpha particle will stop moving.
B) The alpha particle will reverse its direction.
C) The alpha particle will be deflected in a curve path.
Eliminate
D) The alpha particle will continue to travel in a straight line.

Answers

Answer: C) The alpha particle will be deflected in a curve path.

Explanation:  

Alpha particle carries +2 charge. It is a Helium ion: He⁺²

When any charged particle enters a magnetic field, it experiences a magnetic force perpendicular to direction of the velocity of the particle and the direction of the magnetic field. Thus, the alpha particle moves in a circular path.

Hence, the correct answer is: The result of an alpha particle coming into a magnetic field is C) The alpha particle will be deflected in a curve path.

The right answer for the question that is being asked and shown above is that: "C) The alpha particle will be deflected in a curve path. " the result of an alpha particle coming into a magnetic field is that C) The alpha particle will be deflected in a curve path. 

Bands of waves in the electromagnetic spectrum are arranged in order ofa. frequency and wavelength.
b. chronology.
c. discovery.
d. use and applications.

Answers

Bands of waves in the electromagnetic spectrum are arranged in order of frequency and wavelength. The answer is letter A. The rest of the choices do not answer the question above.

[URGENT] A swimmer wants to end up at a dock due north of her starting position on the south shore of a river. In still water her maximum speed is 1.25m/s. The river has a current flowing 0.35m/s [E].a.) calculate the direction that she must aim herself through the water to arrive at the dock.

b.) how long will it take for the swimmer to cross the river, if the distance from the starting point to the dock is 300 m?

Please show all your work, thanks

Answers

a)The direction that she must aim herself through the water to arrive at the dock will be 73.7°.

b)The time it takes for the swimmer to cross the river will be 240 seconds.

What is speed?

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.

Given data;

The maximum speed of the swimmer In still water,v=1.25m/s.

Speed of river = 0.35m/s [E]

α is the direction that she must aim herself through the water to arrive at the dock.

t is the time it takes for the swimmer to cross the river

The distance from the starting point to the dock is,s = 300 m

If the swimmer were to swim straight, the water's speed along the x-axis would have to match that of the swimmer.

\rm v_w = v_x  = 0.35  \ m/sec

α is the angle between vx and v

The value of the

\rm cos \alpha = (v_x)/(v) \n\n \alpha = cos ^(-1)((0.35)/(1.25) )\n\n \alpha = 73.7 ^0

The time that the swimmer will use for reaching the dock;

s = V×t

t = s/(v)

t=300 m / 1.25 m/sec

t = 240 seconds

Hence, the direction that she must aim herself through the water to arrive at the dock. and it takes for the swimmer to cross the river will be 73.7° and 240 seconds respectively.

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Explanation:

As you can see in the picture, we want the swimmer to go on a straight line, so the speed of the water must be equal to the speed of the swimmer along the x-axis. We also know the value of v, so we can calculate the of the cosine of the angle (alpha) between Vx and V. Thanks to the fundamental relation of gioniometry (cos^2(x) + sin^2(x) = 1) we can find the sine of alpha and calculate Vy. With Vy we can calculate the time that the swimmer will use for reaching the dock: s = Vy * t => t = s/(Vy).

I'll let you do all the calculations, you just have to plug in values.