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mote1985 [20]
3 years ago
9

A rare and valuable antique chest is being moved into a truck using a 4.00 m long ramp. the kj weight of the chest plus packing

material is 1,500 n. if the truck bed is 1.00 m above the ground, find the work done by the movers as they slide the chest up the ramp if the coefficient of friction between the chest and the ramp is 0.200.
Physics
1 answer:
Lisa [10]3 years ago
7 0

First let us calculate for the angle of inclination using the sin function,

sin θ = 1 m / 4 m

θ = 14.48°

 

Then we calculate the work done by the movers using the formula:

W = Fnet * d

 

So we must calculate for the value of Fnet first. Fnet is force due to weight minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

 

So the work exerted is equal to:

W = 84.526 N * 4 m

<span>W = 338.10 J</span>

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Treatment water from the plant would affects the communities present in the downstream of the river.

Answer: Option (a)

<u>Explanation:</u>

Wastewater treatment plant is a process used to treat the water that flows from the rivers, streams, and lakes.This plant removes the dust, sand, sediments present in the river water.

When this treated water is released into the downriver stream, it affects the health of aquatic organisms present in that stream.Use of chemicals in treating the wastewater may cause genetic problems to the communities present in that region.

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3 years ago
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The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation
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Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is 920\ x\ 10^{-9} m farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      \vartriangle y =\frac{m \lambda L}{d}

                      m=0,\pm 1,\pm 2,\pm 3,.....

  1. m is the order number.
  2. \lambda is the wavelength of the monochromatic light.
  3. L is the distance between the screen and the two slits.
  4. d is the distance between the slits.
  • Part A:  a) In the above equation for the position of bright fringes we can see that if the wavelength of the light \lambda is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
  • Part B:  b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
  • Part C:  a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
  • Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at  the center of the fringe E in the screen we use the condition for constructive interference. That says that the  path length difference Δr between rays coming from the left and right slit must be \vartriangle r=m \lambda

        We simply replace the values in that equation :

                      \vartriangle r= m \lambda =2.\ 460\ nm

                      \vartriangle r= 920\ x\ 10^{-9} m

         The dot in the center of fringe E is 920\ x\ 10^{-9}m farther from the left slit than from the right slit.

     

       

       

     

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Answer:

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It  is eight times more than the star A.

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luminousity depends upon the two factors are:

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2) The star distance from us

The more distance of an object the dimmer it appears.

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