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svet-max [94.6K]
3 years ago
13

One simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching to

p speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of
Physics
1 answer:
yaroslaw [1]3 years ago
3 0

Complete Question:

One simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching top speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of 11.5 m/s in 2.24 s, what will be his total time?

Answer:

total time = 6.24 s

Explanation:

Using the equation of motion:

v = u + at

initial speed, u = 0 m/s

v = 11.5 m/s

t = 2.24 s

11.5 = 0 + 2.24a

a = 11.5/2.24

a = 5.13 m/s²

For the total time spent by the sprinter:

s = ut + 0.5at²

100 = 0.5 * 5.13 * t²

t² = 100/2.567

t² = 38.957

t = √38.957

t = 6.24 s

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Question #14
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The decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot bind to another atom, which is why hydrogen has a single atom.

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In conclusion, the decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot join another atom, which is why the hydrogen has only one atom.

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6 0
2 years ago
Three point charges, two positive and one negative, each having a magnitude of 20 C are placed at the vertices of an equilateral
Daniel [21]

The resultant force on the positive charge  is mathematically given as

X=40N

<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>

Question Parameters:

Three-point charges, two positive and one negative, each having a magnitude of 20

Generally, the -ve charge   is mathematically given as

Q+=\sqrt{x^2+x^2+2x.xcos120}\\\\Q+=\sqrt{2x^2+2x*(1/2)}

Q+=X

Therefore

x=\frac{Kq1q2}{r2}\\\\x=\frac{9*10^9*20*10^{-6}*20*10^{-6}}{(30*10^-2)^2}

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2 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
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Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
A car speeds up from 12.0 m/s to 16.0 m/s in 8.00s what is the acceleration
Lerok [7]

Answer:

0.5m/s²

Explanation:

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7 0
3 years ago
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