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3 years ago

Write your question here (Keep it simple and clear to get the best answer)what is the dimensions of angular velocity

Physics

1 answer:

frutty [35]3 years ago

7 0

Answer:

The dimensions of angular velocity is $[T^{-1}]$

Explanation:

We know that linear velocity is equal to the rate of change of linear displacement.

Similarly, in rotational motion, the analogous term for linear velocity is angular velocity.

Angular velocity is defined as the rate of change of angular displacement.

The angular displacement is the measure of the angle rotated by an object about a fixed point or the axis of rotation.

Therefore, the unit of angular displacement is measured in radians. We know that, the radian is a dimensionless quantity. So, its dimension is $[M^0L^0T^0]$

Now, time is measured in seconds. So, dimension of time is $[M^0L^0T]$

Therefore, the dimensions of angular velocity is given as:

$Angular\ velocity =\frac{Angular\ displacement}{Time}\\\\Dimensions=\frac{[M^0L^0T^0]}{[M^0L^0T]}\\\\Dimensions =[T^{-1}]$

Therefore, the dimensions of angular velocity is $[T^{-1}]$

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Answer:

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Average velocity is the change in displacement of a body with respect to time.

Velocity = ∆S/∆t

∆S = 100m - 70m

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∆t = 2min - 1 min

∆t = 1min = 60secs

Substitute the given parameters into the formula for velocity

Velocity = 30m/60s

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4 years ago

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Your friend has slipped and fallen. To help her up, you pull with a force F, as the drawing shows. The vertical component of thi

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This question can be solved using the concept of the composition of the vector.

(a) The magnitude of F is b "164.27 N".

(b) The angle of the vector from horizontal is " ".

(a)

From the composition of the rectangular components of a vector, we know that the magnitude of the resultant vector is given by the following formula:

$F = \sqrt{F_x^2+F_y^2}$

where,

$F_x$ = x-component of the force = 122 N

$F_y$ = y-component of the force = 110 N

Therefore,

$F = \sqrt{(122\ N)^2+(110\ N)^2}$

F = 164.27 N

(b)

The angle of the vector is given by the following formula:

$\theta = tan^{-1}(\frac{F_y}{F_x})\\\\\theta = tan^{-1}(\frac{110\ N}{122\ N})\\\\$

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Learn more about the composition of vector here:

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3 years ago

A projectile is fired with an initial speed of 37.6 m/s at an angle of 43.6° above the horizontal on a long flat firing range. P

Olenka [21]

Answer:

A) The maximum height reached by the projectile is 34.3 m.

B) The total time in the air is 5.29 s.

C) The range of the projectile is 144 m.

D) The speed of the projectile 1.80 s after firing is 28.4 m/s.

Explanation:

Please, see the attached figure for a better understanding of the problem.

The position and velocity vectors of the projectile at time "t" are as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = vector position at time t

Let´s place the origin of the frame of reference at the launching point so that x0 and y0 = 0.

A) At the maximum height, the vertical component of the velocity is 0 (see figure). Then, using the equation for the y-component of the velocity vector, we can obtain the time at which the projectile is at its maximum height:

vy = v0 · sin α + g · t

0 = 37.6 m/s · sin 43.6° - 9.8 m/s² · t

- 37.6 m/s · sin 43.6° / -9.8 m/s² = t

t = 2.65 s

The height of the projectile at this time will be the maximum height. Then, using the equation of the y-component of the vector position:

y = y0 + v0 · t · sin α + 1/2 · g · t² (y0 = 0)

y = 37.6 m/s · 2.65 s · sin 43.6° - 1/2 · 9.8 m/s² · (2.65)²

y = 34.3 m

The maximum height reached by the projectile is 34.3 m.

B) When the projectile reaches the ground, the y-component of the position vector is 0 (see vector "r final" in the figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 37.6 m/s · t · sin 43.6° - 1/2 · 9.8 m/s² · t²

0 = t · (37.6 m/s · sin 43.6° - 1/2 · 9.8 m/s² · t) (t = 0, the initial point)

0 = 37.6 m/s · sin 43.6° - 1/2 · 9.8 m/s² · t

- 37.6 m/s · sin 43.6° /- 1/2 · 9.8 m/s² = t

t = 5.29 s

The total time in the air is 5.29 s.

C) Having the total time in the air, we can calculate the x-component of the vector "r final" (see figure) to obtain the horizontal distance traveled by the projectile:

x = x0 + v0 · t · cos α

x = 0 m + 37.6 m/s · 5.29 s · cos 43.6°

x = 144 m

The range of the projectile is 144 m.

D) Let´s find the velocity vector at that time:

v = (v0 · cos α, v0 · sin α + g · t)

vx = v0 · cos α

vx = 37.6 m/s · cos 43.6°

vx = 27.2 m/s

vy = v0 · sin α + g · t

vy = 37.6 m/s · sin 43.6° - 9.8 m/s² · 1.80 s

vy = 8.29 m/s

Then, the vector velocity at t = 1.80 s will be:

v = (27.2 m/s, 8.29 m/s)

The speed is the magnitude of the velocity vector:

$|v| =\sqrt{(27.2 m/s)^{2} +(8.29 m/s)^{2}} = 28.4 m/s$

The speed of the projectile 1.80 s after firing is 28.4 m/s.

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3 years ago