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notka56 [123]
3 years ago
13

An athlete has a final velocity of 9.5 m/s. If the athlete starts from the blocks and has an initial velocity of 0 m/s, what is

the athlete’s acceleration if he runs the race as 41s?
Physics
1 answer:
arlik [135]3 years ago
4 0
Answer :

Explanation:
a =vf -vi/t
a = 9.5m/s -om/s/41 s
a = 0. 23 m/s²
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A

Explanation:

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If a plane is inclined at an angle 5o with the horizontal and a cylindrical metal is allowed to roll between two fixed height 92
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0.61 second.

Explanation:

So, from the question we are given the following useful parameters or data or information that is going to assist us in solving this question and they are;

(1). Inclination of plane, θ = 50°.

(2). ''The horizontal and a cylindrical metal is allowed to roll between two fixed height 92cm apart on the plane" that is height = 0.92 m(92cm).

(3). The "acceleration of the cylinder to be 2/3 × g Sin (theta) and g = Acceleration due to gravity = 9.8m/s^2

Soz let us delve right into the solution to the question above. We will be making use of the mathematical representation or Equation below;

s = (1/2) a × t².

Where s = height, a = acceleration due to gravity and t = time.

0.92 = (9.8 × sin 50°/ 3) × t².

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t = √ 0.37.

t =0.61 seconds.

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Quia<br> Active<br> 16:05:5<br> A plant root is an example of
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q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

For the circular motion of electron

qBr = mv

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v = 2.8 x 10⁴ m/s

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v' = linear speed

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t = time taken = \frac{2\pi m}{qB} = \frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{^{-19}})(80\times 10^{-6})} = 4.5 x 10⁻⁷ sec

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