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kaheart [24]
3 years ago
6

Two runners start simultaneously from the same point on a circular 200-m track and run in the same direction. One runs at a cons

tant speed of 6.20 m/s and the other runs at a constant speed of 5.50 m/s.When will the fast one first overtake ("lap") the slower one?How far from the starting point will each have run at that instant?When will the fast one overtake the slower one for the second time?How far from the starting point will they be at that instant?
Physics
1 answer:
sladkih [1.3K]3 years ago
5 0

Answer

given,

length of the track = 200 m

speed of runner 1 = 6.20 m/s

speed of the runner 2 = 5.5 m/s

distance covered by the fastest runner, d = 6.2 t

distance covered by the slow runner,d = 5.50 t + 200

time taken to overtake for the first time

 6.2 t = 5.50 t + 200

  0.7 t = 200

       t = 285.7 s

distance covered from the starting point

 d = 6.2 t

 d = 6.2 x 285.7

 d = 1771.43 m

for next overtaking time taken will be double

 t' = 258.7 x 2 = 571.4 s

distance travel by the runner from the starting point

d' = 6.2 t'

d' = 6.2 x 571.4

d' = 3542.68 m

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☁️ Answer ☁️

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7 0
2 years ago
Read 2 more answers
Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds
melisa1 [442]

Answer:

120 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 12 m/s²

t = 10 s

Find: v

v = at + v₀

v = (12 m/s²) (10 s) + 0 m/s

v = 120 m/s

6 0
3 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
3 years ago
A car weighing 14,700 N is speeding down a highway with a velocity of 99 km/h. What is the
tankabanditka [31]

Answer: 148348.6239 kg•m/s

Explanation: Firstly, we need to convert the 14700 N into kilograms, and to do so, use the formula net force is equal to mass times acceleration and rearrange the formula to find mass like shown below...

F = ma

F/a = m

14700/9.81 = 1498.470948 kg, this is your mass

Now that we convert it into kilograms, plug all the numbers into the variable of the momentum formula.

Momentum formula is P = mass x velocity

Like this:

P = 1498.470948 x 99

p = 148348.6239 kg•m/s.

I believe that is your answer, hope that helps you even a bit out.

Thanks.  

7 0
2 years ago
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