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shepuryov [24]
3 years ago
6

Please look at the photo and give me the answer :)))

Physics
1 answer:
san4es73 [151]3 years ago
5 0

Answer:

c

Explanation:

efficiency is measured in % hence 0.02% is smaller than 20%

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If a rock falls from a cliff, at what point are its kinetic energy and its potential energy the same? Ignore air resistance.
nekit [7.7K]

Answer:

The gravitational potential energy it had from being above the ground is converted to kinetic energy as the rock falls. As kinetic energy increases, the velocity of the rock will also increase.

Explanation:

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What property of an object can you determine using mass and volume?
Eddi Din [679]
The answer is density 
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3 years ago
What is the speed of a transverse wave in a rope of length 2.00 m and mass 60.0 g under a tension of 500 n?
drek231 [11]

The formula we can use in this case would be:

v = sqrt (T / (m / l))

Where,

v = is the velocity of the transverse wave = unknown (?)

T = is the tension on the rope = 500 N

m = is the mass of the rope = 60.0 g = 0.06 kg

 l = is the length of the rope = 2.00 m

Substituting the given values into the equation to search for the speed v:
v = sqrt (500 N/(0.06 kg /2 m)) 
v = sqrt (500 * 2 / 0.06) 
v = sqrt (16,666.67) 
<span>v = 129.10 m/s</span>

7 0
3 years ago
Calculate the electric field at the location of a charge if the electric force is 10 newtons and the charge on the particle is 6
kolezko [41]

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6 0
3 years ago
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As
Naddik [55]

Answer:

a

   \lambda = 1.667 nm

b

     \theta  =  0.8681^o

Explanation:

From the question we are told that

   The distance of separation is d  =  0.220 \ mm  =  0.00022 \ m

    The  is distance of the screen from the slit is  D   =  2.60 \ m

    The distance between the central bright fringe and either of the adjacent bright   y  =  1.97 cm  =  1.97 *10^{-2}\ m

Generally  the condition for constructive interference is  

      d sin \tha(\theta ) =  n \lambda

From the question we are told that small-angle approximation is valid here.

So    sin (\theta ) = \theta

=>        d \theta  =  n \lambda

=>        \theta =  \frac{n *  \lambda }{d }

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         y  =  D * sin (\theta )

From the question we are told that small-angle approximation is valid here.

So

       y  =  D * \theta

=>   \theta  =  \frac{ y}{D}

So

     \frac{n *  \lambda }{d } = \frac{y}{D}

     \lambda =\frac{d * y }{n * D}

substituting values

       \lambda =  \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }

        \lambda = 1.667 *10^{-6}

        \lambda = 1.667 nm

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     dsin (\theta ) =  n \lambda

    \theta  =  sin^{-1}[\frac{ n  *  \lambda }{d} ]

    \theta  =  sin^{-1}[\frac{ 2  *  1667 *10^{-9}}{ 0.00022} ]

    \theta  =  0.8681^o

7 0
4 years ago
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