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Afina-wow [57]
1 year ago
5

Velocity vector and acceleration vector in a uniform circular motion are related as.

Physics
1 answer:
mr_godi [17]1 year ago
8 0

They are related as \bold{\underline {v}\,.\,\underline a }= \bold{0}

  • In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
  • This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
  • For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
  • Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
  • In mathematics, 2 vectors (\underline p , \underline q) that are perpendicular to each other have a quality that their dot product (\underline p\,.\, \underline q) equal to zero vector (\bold 0) which is written as \undeline p\,.\, \underline q = \bold 0.
  • This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner \underline v\,.\, \underline a =\bold 0.

#SPJ4

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In a double-slit arrangement, the slits are separated by a distance equal to 100 times the wavelength of the light passing throu
VladimirAG [237]

Solution :

The conditions for the maximum in the Young's experiment is :

d sin θ = m λ,     where m = 0, 1, 2, 3, .....

The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,

d sin θ =  λ

$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$

Given : d = 100 λ

$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$

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6 0
3 years ago
A van traveling down a slope with a uniform acceleration of 2.15 meters/second2 attains a speed of 20.00 meters/second after 7.0
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6 0
4 years ago
Read 2 more answers
A navy diver hears the underwater sound wave from an exploding ship across the harbor. They immediately lift their head out of t
tamaranim1 [39]

Answer:

s = 1800 m = 1.8 km

Explanation:

The distance, the speed, and the time of reach of the sound are related by the following formula:

s = vt

where,

s = distance

v = speed

t = time

FOR WATER:

s = v_wt ---------------------- eq (1)

where,

s = distance between ship and diver = ?

v_w = speed of sound in water = 1440 m/s

t = time taken by sound in water

FOR AIR:

s = v_a(t+4\ s) ---------------------- eq (2)

where,

s = distance between ship and diver = ?

v_a = speed of sound in water = 344 m/s

t + 4 s = time taken by sound in water

Comparing eq (1) and eq (2),because distance remains constant:

v_wt=v_a(t+4\ s)\\\\(1440\ m/s)t = (344\ m/s)(t+4\ s)\\(1440\ m/s - 344\ m/s)t=1376\ m\\t = \frac{1376\ m}{1096\ m/s}

t = 1.25 s

Now using this value in eq (1):

s = (1440\ m/s)(1.25\ s)

<u>s = 1800 m = 1.8 km</u>

4 0
3 years ago
in general, what happens to the energy, order, and spacing of particles when a solid turns to a liquid?
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