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solong [7]
4 years ago
12

if a bicyclist, with initial velocity of zero, steadily gain speed until reaching a final velocity of 26m/s, how far away did sh

e travel during the race (in the same amount of time?
Physics
1 answer:
Alja [10]4 years ago
5 0

Answer:

<h2><em>34.46m</em></h2>

Explanation:

Using one of the equation of motion to solve the question. According to the equation v² =u²+2as where;

v is the final velocity of the  bicyclist = 26m/s

u is the initial velocity of the  bicyclist = 0m/s

a is the acceleration due to gravity = 9.81m/s

s is the distance covered during travel

Substitute the given parameters into the formula above to get the distance traveled

26² = 0² + 2(9.81)s

676 = 19.62s

Divide both sides  by 19.62

676/19.62 = 19.62s/19.62

s = 34.46m

<em>The distance traveled by the  bicyclist during the race is 34.46m</em>

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Answer:

200000J

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6 0
3 years ago
What is the anomalous expansivity of water
lora16 [44]

Answer:

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6 0
3 years ago
A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the
ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

8 0
3 years ago
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4 0
3 years ago
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NikAS [45]

Answer:

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3 0
3 years ago
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