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nadya68 [22]
3 years ago
10

Two runners start a race. After 2 seconds, they both have the same velocity. If they both started at the same time, how do their

average accelerations compare?
Physics
2 answers:
djyliett [7]3 years ago
7 0

let each runner starts from rest

v₀ = initial velocity of each runner = 0 m/s

t = time of travel = 2 sec =

a = average acceleration

average acceleration is given as

a = (v - v₀)/t

inserting the values

a =  (v - 0)/t

a = v/t

since each runner gains equal amount of speed in given time interval of "t". hence the average acceleration of each runner is same.

arsen [322]3 years ago
3 0

Answer:

Two runners start a race. After 2 seconds, they both have the same velocity.  they both started at the same time, how do their average velocities compare? They have the same acceleration rate.

Explanation:

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Answer:

  F = -49.1   10³ N

Explanation:

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    x = 11.00 cm (1 m / 100 cm) = 0.110 m

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   a = -1320²/(2 0.110)

   a = -7.92 10⁶ m / s²

With Newton's second law we find the force

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3 years ago
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sleet_krkn [62]

The electric potential V(z) on the z-axis is :  V = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z  

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq =  б( 2πR ) dr

dV  \frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = \int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,

 V = \pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]

     = \pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} }  -z ]

Therefore the electric potential V(z) = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

Also

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

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