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aliina [53]
2 years ago
7

Definition: - When a body is moving in circular path at a distance r from its center, this is its velocity at any instant?

Physics
2 answers:
Natali [406]2 years ago
4 0

Answer:

tangential velocity

Explanation:

kupik [55]2 years ago
3 0
Tangential velocity!
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A small particle starts from rest from the origin of an xy-coordinate system and travels in the xy-plane. Its acceleration in th
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Answer:

The x-coordinate of the particle is 24 m.

Explanation:

In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion

Xf=Xo+Voxt+0.5axt²(I)

Yf=Yo+Voyt+0.5ayt² (II)

Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.

The particle starts from rest from the origin, therefore:

Vox=Voy=0

Xo=Yo=0

Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:

12=0+(0)t+ 0.5(1.0)t²

12=0.5t²

Dividing by 0.5 and extracting thr squareroot both sides:

t=√12/0.5

t=√24 = 2√6

Replacing t=2√6, ax=2.0,Xo=0 and Vox=0 in (I) to obain the x-coordinate:

Xf=0+0t+0.5(2.0)(2√6)²

Xf= 24 m

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Find deacceleration An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When
earnstyle [38]

Answer:

The deceleration is  a = -0.7273 \ m/s^2  

Explanation:

From the question we are told that

  The distance of the car from the crossing is d = 360 \  m

   The speed is u  = 16 \  m/s

    The reaction time of the engineer is  t = 0.53 \ s

Generally the distance covered during the reaction time is  

      d_r = u * t

=>   d_r = 16 * 0.53

=>   d_r = 8.48 \ m

Generally distance of the car from the crossing after the  engineer reacts is

      D = d- d_r        

=> D = 360 - 8.48      

=> D = 352 \ m

Generally from kinematic equation

      v^2 = u^2 + 2as

Here v is the final velocity of the car which is  0 m/s

So

        0^2 = 16^2 + 2 * a * 352

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