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Talja [164]
3 years ago
5

HERES THE ANSWER AND QUESTION

Physics
2 answers:
tatyana61 [14]3 years ago
5 0

Answer:

agree

Explanation:

kobusy [5.1K]3 years ago
5 0

Answer:

THE ANSWER IS A,B,F

Explanation:

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Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic
Anni [7]

Answer:

force for start moving is 7.49 N

force for moving constant velocity 2.25 N

Explanation:

given data

mass = 7.65 kg

kinetic coefficient of friction = 0.030

static coefficient of friction = 0.10

solution

we get here first weight of block of ice that is

weight of block of ice = mass  ×  g

weight of block of ice = 7.65 × 9.8 = 74.97 N

so here Ff = Fa

so for force for start moving is

Fa = weight × static coefficient of friction  

Fa = 74.97 × 0.10

Fa =  7.49 N

and

force for moving constant velocity is

Fa =  weight × kinetic coefficient of friction

Fa = 74.97 × 0.030

Fa = 2.25 N

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In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between
kap26 [50]

The separation between the slits is d = 8.96

What is fringe width?

  • Fringe width is the distance between two consecutive bright spots (maximas, where constructive interference take place)
  • Or two consecutive dark spots (minimas, where destructive interference take place).

Fringe width is given by β = λL/d

In the first case fringe width is β1 = λLA /d   = 625 x 10-9 x 0.36 / ( 1.4 x 10-5 )  = 0.016071428 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.016071428  = 6.222  

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 6.

In the second case fringe width is β1 = λLAB /d   = 625 x 10-9 x 0.25 / ( 1.4 x 10-5 )  = 0.011160714 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.011160714  = 8.96

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 8.  The ninth one will not be seen since the screen is less a little less in width.

Learn more about fringe width

brainly.com/question/14438105

#SPJ4

<u>The complete question is -</u>

In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between the slits is d = 1.4 × 10-3 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.36 m, whereas in another version it is LB = 0.25 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting. --Tm = 3 (Bright fringe) ++m = 0 (Bright fringe) -m = 3 (Bright fringe) Figure 27.7

5 0
1 year ago
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