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Allushta [10]
3 years ago
13

Calculate the speed of an object which Travels 30 km distance in 4 hours​

Physics
1 answer:
nexus9112 [7]3 years ago
7 0

Answer:

The object is traveling 7.5 km per hour.

Explanation:

30/4 = 7.5

To check do 7.5 * 4 and you get 30

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How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 1.6 % increase
morpeh [17]

Answer:

The value is  v_o =  5.488 \  m/s

Explanation:

From the question we are told that

     The emitted frequency increased by \Delta f =  1.6 \% = 0.016 \

Let assume that the initial value of the emitted frequency is

      f =   100\%  =  1

Hence new frequency will be  f_n  =  1 +  0.016 = 1.016

Generally from Doppler shift equation we have that

         f_1 =  [\frac{ v \pm v_o}{v \pm + v_s } ] f

Here v  is the speed of sound with value  v =  343 \ m/s

         v_s is the velocity of the sound source which is v_s = 0 \ m/s because it started from rest

         v_o  is the observer velocity So

Generally given that the observer id moving towards the source, the Doppler frequency becomes

                   f_1 =  [\frac{ v + v_o}{v + 0 } ] f

=>                1.016  =  [\frac{ 343  + v_o}{343 } ] * 1  

=>                v_o =  5.488 \  m/s

         

8 0
3 years ago
ou are out stargazing with your 13.4-cm telescope. You point your telescope at an interesting formation in the sky, which you th
Alinara [238K]

Answer:

θ = 4.716 10⁻⁶ rad

Explanation:

In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.

We use the diffraction equation for a slit

            a sin θ = m λ

The minimum occurs at m = 1

             sin θ = λ / a

Since the angles in these systems are very small, we can approximate the sine to its angle in radians

             θ = λ / a

The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number

           θ = 1.22 λ / a

Let's calculate

          θ = 1.22 518 10⁻⁹ / 13.4 10⁻²

          θ = 4.716 10⁻⁶ rad

8 0
3 years ago
If the orbital distance of earth was greater than its actual value, then
KiRa [710]

Answer:

it Farther then it seems

Explanation:cause it value

4 0
3 years ago
Light from a sodium lamp 1l = 589 nm2 illuminates two narrow slits. the fringe spacing on a screen 150 cm behind the slits is 4.
BartSMP [9]

d = 0.221 mm is the spacing (in mm) between the two slits..

here,

wavelength  = 589 nm

D = 150 cm  =  1.5 m

y  = 4 mm

then ,

y = mλD/d

for m  = `1

0.004  = 589*10^-9 * 1.5/d

d = 2.21*10^-4 m  

d = 0.221 mm

slit spacing is 0.221 mm

The sesmic characteristics of a train depend on the speed of the train, the distance between the wheels, and the hardness of the ground beneath the rails.

Philadelphia attack efficiency aims to improve spacing It continues to lag despite offseason moves, and Harden could have single-handedly fixed these issues.

In a typical hydrogen bond, as atoms move further up the energy ladder, The distance between the energy levels decreases. For Scientists Working at Such Small Scales, these “charging times” are likely candidates for the cause of the distance effect.

Learn more about spacing here:

brainly.com/question/15862947

#SPJ4

7 0
2 years ago
How are a wave's energy and the wave's amplitude related?
dem82 [27]
Answer:
Energy is proportional to the square of the amplitude

Explanation:
The energy of a certain wave is defined using its magnitude.
The two quantities are related directly. This means that as the amplitude of the wave increases, its energy increases and vice versa.

Energy is directly proportional to the square of the magnitude of the wave. This means that:
If we have new amplitude = 2 * old amplitude
We will have new energy = (2)² * old energy = 4 * old energy

Hope this helps :)
7 0
3 years ago
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