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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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Help me for a physics project please
lesantik [10]

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-sound travels 4.3 times faster in water than air

-sounds are waves that pass through our ears via vibrations and travel by vibrations of molecules

-different types of sound like audible, inaudible, infrasonic, ultrasonic,

-sounds waves are either longitudinal, mechanical and pressure waves

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4 0
3 years ago
A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 10^3 N/C. If th
tensa zangetsu [6.8K]

Answer:

The net charge is 67.89 \mu C

Solution:

As per the question:

Mass of the plastic bag, m = 12.0 g = 12.0\times 10^{-3}\ kg

Magnitude of electric field, E = 10^{3}\ N/C

Angle made by the string, \theta = 30^{\circ}

Now,

To calculate the net charge, Q on the ball:

Vertical component of the tension in the string, T = Tcos\theta

Horizontal component of the tension in the string, T = Tsin\theta

Now,

Balancing the forces in the x-direction:

Tsin\theta = QE

Q = {Tsin\theta}{E}                             (1)

Balancing the forces in the y-direction:

Tcos\theta = mg

where

g = acceleration due to gravity = 9.8\ m/s^{2}

Thus

T = \frac{mg}{cos\theta }

T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N

Use T = 0.1357 N in eqn (1):

Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C

Q = 67.89\times 10^{- 5}\ C = 67.89\mu C

7 0
3 years ago
The Moon's appearance as seen from Earth during the 28-day lunar cycle. You can see how the changes happen by looking at the Sun
Pie

Answer:     D

Explanation:

5 0
3 years ago
Read 2 more answers
Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
WITCHER [35]

Answer:

Explanation:

a )

If it is totally absorbed pressure is calculated as follows .

Pressure = I / c where I is intensity of light falling .

= 1000 / 3 x 10⁸

= 3.33 x 10⁻⁶ N / m²

b ) weight of tritium atom

=  3 x 1.67 x 10⁻²⁷ kg

acceleration = force / mass

=    3.33x 10⁻⁶ / 3 x 1.67 x 10⁻²⁷

= .6646 x 10²¹ m /s²

= 66.46 x 10¹⁹ m / s²

7 0
3 years ago
Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at
Stolb23 [73]

Answer:

a. Fx = -8.089 N b. Fy = 3.525 N c. 8.824 N d. 336.45°

Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C  

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

7 0
3 years ago
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