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Stells [14]
2 years ago
9

In the video, light passing through two slits creates a pattern of bright and dark fringes on a screen. The bright fringes resul

t from __________; the dark fringes result from __________. In the video, light passing through two slits creates a pattern of bright and dark fringes on a screen. The bright fringes result from __________; the dark fringes result from __________. Diffraction, constructive interference Diffraction, destructive interference Constructive interference, destructive interference Diffraction, interference
Physics
1 answer:
stealth61 [152]2 years ago
4 0

Answer:

Constructive interference: destructive interference

Explanation:

Constructive interference occurs when waves are in phase or sync with one another and will lead to brighter and bolder results while destructive interference occurs when waves are out of phase with one another.

In this scenario, light passing through two slits creates a pattern of bright and dark fringes on a screen. The bright fringes result from constructive interference ; the dark fringes result from destructive interference.

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A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri
Alla [95]

Answer:

q=1\times10^{-8}C

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

F=\frac{KQq}{d^{2}}

By substituting the values

0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}

q=1\times10^{-8}C

Thus, the charge on the ball bearing is q=1\times10^{-8}C

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2 years ago
Please help calculate them<br> A-f please very urganr
valentina_108 [34]
Three 40w lamp for 6 hours
8 0
3 years ago
A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?
anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\  \end{gathered}

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}

now, replace

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}

hence, the answer is 0.495 m/s

3 0
11 months ago
Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time
anzhelika [568]

The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let \overset \rightarrow {v_y} represent the velocity of the boat across the river in the direction, \overset \rightarrow {d_y}, and let, \overset \rightarrow {v_x}, represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = \overset \rightarrow {v_y}

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is \overset \rightarrow {v_y}, such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

\overset \rightarrow {v_y} = 3 m/s

\overset \rightarrow {v_x} = 4 m/s

The width of the river, w_y = 6 meters, we have;

  • The resultant velocity = \sqrt{(3 \ m/s)^2 + (4 \ m/s)^2} =5 \ m/s

The direction, θ, is given as follows;

\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}

The length of the path of the boat, <em>l</em>, is given as follows;

l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = \dfrac{l}{v}, therefore;

  • t = \dfrac{10}{5} = 2
  • The time it takes to cross the river, t = 2 seconds

Considering only the y-components, we have;

t = \dfrac{w_y}{v_y}

Therefore;

t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s

Which expresses that the time taken is the same and given that the

vectors of the velocities of the river and the boat are perpendicular, the

distance covered in the direction of the boat is unaffected by the velocity

of the river.

Learn more vectors here:

brainly.com/question/15907242

5 0
3 years ago
What is the order of the events for the water cycle on a typical warm day?
Rudik [331]
B precipitation,condensation,precipitation
6 0
2 years ago
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