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4 years ago

Which is the correct formula for finding the frequency of an electromagnetic wave? f = c – f = f = + c f =

Physics

1 answer:

liubo4ka [24]4 years ago

6 0

The answer for the online test is going to be the last one, D

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How much force is required to move a sled 5 meters if a person uses 60 J of work?

Brut [27]

W=F*S

W - Work
F - Force
S - Distance (from latin word 'spatium)

so...
S= 5 (m)
W=60 (J)

W=F*S
F=W/S
F=60/5=12 J/m = 12 N (Newtons)

6 0

3 years ago

What is the simulation theory?

Inessa05 [86]

The simulation hypothesis contends that reality is in fact a simulation (most likely a computer simulation), of which we, the simulants, are totally unaware. Some versions rely on the development of simulated reality, a fictional technology.

7 0

3 years ago

A block of wood 3 cm on each side has a mass of 27 g. what is the density of the block

butalik [34]

Density: Mass/Volume

Volume: 3 x 3 x 3cm
= 27cm

27/27
= 1g/cm^3

7 0

3 years ago

Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation

gogolik [260]

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

3 0

3 years ago

In Young's double slit experiment, 427 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What

anastassius [24]

Answer:

λ' = 379.22nm

Explanation:

In order to find the longest wavelength that allows one to dark fringe coincides with the bright fringe, you use the following formulas:

$y_{bright}=\frac{m\lambda D}{d}$ (1)

$y_{dark}=(m+\frac{1}{2})\frac{\lambda' D}{d}$ (2)

y-dark and y-bright are the positions of dark fringes and bright fringes respectively.

m: order of the fringe (dark or bright) = 4

D: distance to the screen

d: distance between slits

λ: light for y-bright= 427nm

λ': second light for y-dark = ?

By the information of the statement you know that y-bright = y-dark.

You divide equation (2) into the equation (1) and solve for λ':

$\frac{y_{dark}}{y_{bright}}=1=\frac{(m+1/2)\lambda'}{m\lambda}\\\\\lambda'=\frac{m\lambda}{m+1/2}$ (3)

Finally, you solve the equation (3) by replacing the values of m and the wavelength:

$\lambda'=\frac{4(427nm)}{4+1/2}=379.55nm$

The longest wavelength which produces the fourth dark fringe in the same location for the fourth bright fringe of the first wavelength is 379.22nm

4 0

4 years ago