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OLEGan [10]
3 years ago
11

Please help me with this question

Physics
1 answer:
valkas [14]3 years ago
4 0

Answer: m∠P ≈ 46,42°

because using the law of sines in ΔPQR

=> sin 75°/ 4 = sin P/3

so ur friend is wrong due to confusion between edges

+) we have: sin 75°/4 = sin P/3

=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16

=> m∠P ≈ 46,42°

Explanation:

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In a two-slit experiment using coherent light, the distance between the slits and the screen is 1.10 m, and the distance between
Paul [167]

Answer:

D) 763 nm

Explanation:

Calculation for the wavelength of light

Using this formula

Wavelength of light=Delta Y*Distance / Length

Where,

Delta Y represent the 2nd order bright fringe

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Distance represent the Distance between the slits

Let note that cm to m = (4.2) x 10^-2 and mm to m= ( 0.0400x 10^-3)

Now Let plug in the formula

Wavelength of light=[(4.2 x 10^-2m)(0.0400 x 10^-3m) / 2(1.1m)]*10^-7 meters

Wavelength of light=[(0.042m) (0.0004m)/2.2m]*10^-7 meters

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3 0
2 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
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