All categories

3 years ago

When two light waves arrive at the same place at the same time they create a?

Physics

1 answer:

Montano1993 [528]3 years ago

3 0

Hey there,
<span>They interfere essentially like any other form of wave.
</span>
Hope this helps :))

~Top

You might be interested in

Ice has a density of 0.920 g/cm3. An ice cubes mass is 100 g. What is its volume?

Tasya [4]

Answer:

$\boxed {\tt 108.695652 \ cm^3}$

Explanation:

Volume can be found by dividing the mass by the density.

$v=\frac{m}{d}$

The mass is 100 grams and density is 0.920 grams per cubic centimeters.

Therefore,

$m=100 \ g \\d= 0.920 \ g/cm^3$

Substitute the values into the formula.

$v=\frac{100 \ g}{0.920 \ g/cm^3}$

Divide. Note that the grams, or "g" will cancel each other out.

$v- 108.695652 \ cm^3$

The volume of the ice cube is 108.697652 cubic centimeters.

3 0

3 years ago

Does anyone want to do my science home work lol

USPshnik [31]

I barely do my own homework lol

6 0

2 years ago

Read 2 more answers

Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of

Fudgin [204]

Answer:

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

Explanation:

let 'd' be distance b/w Los Angeles and San Francisco i.e 400 mi

considering ,

Alan's speed $v_A$ =50mph

Beth's speed $v_B$ =60mph

->For Alan:

The time required $t_A$ = d/ $v_A$ = 400/50 => 8h

-> For beth:

The time required $t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h$ => 6h 40m

Alan will reach at 8:00 a.m +8h = 4:00p.m.

Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

3 0

3 years ago

A football player runs from his own goal line to the opposing team's goal line, returning to his forty-yard line, all in 22.4 s.

Bumek [7]

I assume L=120 yards as the length of the football field.

1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
$S=120y+(120-40)y=120y+80y=200 y$
And the time taken is t=22.4 s, so the average speed of the player is
$v= \frac{S}{t}= \frac{200 y}{22.4 s}=8.93 y/s$

2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
$S=120 y-80 y=40 y$
and so, the average velocity is
$v= \frac{S}{t}= \frac{40 y}{22.4 s}= 1.79 y/s$

6 0

3 years ago

Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed

schepotkina [342]

Answer:

12.35m

Explanation:

Hello! To solve this problem we must consider the following:

1. The car moves with constant speed, which means that the distance traveled is equal to the multiplication of time by speed.

X = VT

we solve the equation for time

$T=\frac{x}{V} =\frac{27}{17} =1.588s$

2. The bolt moves with constant acceleration, with acceleration of 9.81m / s ^ 2, so we could apply the following equation.

note=remember that "a uniformly accelerated motion", means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

$Y= VoT-\frac{1}{2}gt^{2}$

where

Vo = Initial speed

T = time

=1.588s

g =gravity=9.8m/s^2

Y = bridge height

solving

$Y= \frac{1}{2}gt^{2}\\Y=(0.5)(9.8)(1.588^2)=12.35m$

4 0

3 years ago