If mass doubled and the force remained constant then the acceleration would also double
Hopes this helps
The magnitude of acceleration is (change in speed) / (time for the change).
Change in speed = (speed at the end) - (speed at the beginning) =
(16 m/s) - (0) = 16 m/s .
Time for the change = 4 s .
Magnitude of acceleration = (16 m/s) / (4 s) = 4 m/s per sec = 4 m/s² .
Answer:
Explanation:
b ) The problem is based on Doppler's effect of sound
f = f₀ x (V - v₀) /(
)
f is apparent frequency ,f₀ is real frequency , V is velocity of sound , v₀ is velocity of observer going away ,
is velocity of source going away
778 = 840 x (340 - 14)/ (340 +
)
340 +
= 341.18
= 1.18 m /s
it will go away from the observer or the cyclist.
speed of train = 1.18 m /s
a )
For a stationary observer v₀ = 0
f = f₀ x V /(
)
= 840 x 340 / (340 + 1.180)
= 837 Hz
Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
From the solution that I have done, the wavelength in the question that we have is 31.88 cm
<h3>How to solve for the wavelength</h3>
The frequency in the question is given as 40/30 = 1.33 hz
Next we have to solve for V
= 425/10
= 42.5 cm/s
v = frequency * wavelength
we have to put in the values in the formula. This would be
42.5 = 1.33 x wavelength
we have to divide through by 1.33 to get the wavelength. This would be
42.5/1.333 = wavelength
31.88 cm = wavelength
Hence we can say that the wavelength in the question that we have here is 31.88 cm
Read more on wavelength here:
brainly.com/question/10728818
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