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Ede4ka [16]
3 years ago
15

4.

Physics
1 answer:
mixer [17]3 years ago
5 0

Answer:

150 J

Explanation:

here, m= 30kg

height = 0.5 m

so

potential energy = mgh

= 30×10×0.5= 30×5

= 150 J

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From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ
V125BC [204]

Answer:

55.8 m/s

Explanation:

f = Actual frequency of sound emitted by car

f' = frequency observed when the car moves = (0.86)f

V = Speed of sound = 343 m/s

v = Speed of car

Frequency observed is given as

f' = \frac{Vf}{V+v}

(0.86) = \frac{(343)}{343 + v}\\(0.86) = \frac{(343)}{343 + v}\\294.98 + (0.86) v = 343 \\v = 55.8 ms^{-1}

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WILL GIVE BRAINLIEST AND 30 POINTS!
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Answer:

The answer is 16 and 24

Explanation:

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Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (
lesya [120]

Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

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3 years ago
A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu
Montano1993 [528]

Answer:

The angle of separation is  \Delta \theta =  0.93 ^o

Explanation:

From the question we are told that

    The angle of incidence is  \theta  _ i  = 56^o

     The refractive index of violet light  in diamond  is  n_v = 2.46

       The refractive index of red light in diamond is n_r = 2.41

      The wavelength of violet light is  \lambda _v = 400nm = 400*10^{-9}m

         The wavelength of red  light is  \lambda _r = 700nm = 700*10^{-9}m

Snell's  Law can be represented mathematically as

         \frac{sin \theta_i}{sin \theta_r} = n

Where \theta_r is the angle of refraction

=>       sin \theta_r  =   \frac{sin \theta_i}{n}

Now considering violet light

               sin \theta_r__{v}}  =   \frac{sin \theta_i}{n_v}

substituting values

                sin \theta_r__{v}}  =   \frac{sin (56)}{2.46}

                 sin \theta_r__{v}}  =  0.337

                 \theta_r__{v}}  =  sin ^{-1} (0.337)

                 \theta_r__{v}}  =  19.69^o

Now considering red light

               sin \theta_r__{R}}  =   \frac{sin \theta_i}{n_r}

substituting values

                sin \theta_r__{R}}  =   \frac{sin (56)}{2.41}

                 sin \theta_r__{R}}  =  0.344

                 \theta_r__{R}}  =  sin ^{-1} (0.344)

                 \theta_r__{R}}  = 20.12^o

The angle of separation between the red light and the violet light is mathematically evaluated as

                  \Delta \theta = \theta_r__{R}} -  \theta_r__{V}}

substituting values

                  \Delta \theta =20.12 - 19.69

                  \Delta \theta =  0.93 ^o

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