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3 years ago

A runner completes a 10 km run in about 45 minutes. What was the runner’s speed in km/h?

Physics

1 answer:

OlgaM077 [116]3 years ago

5 0

Answer: 12.5 km/s

I don't really know how to explain this, but here is your answer.

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6) A boat crosses a river at a constant engine speed of 2.0 m/s under pointed Directly west. The river runs directly south at 2.

TEA [102]

Answer:ligma ;)

Explanation:

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2 years ago

How frequently do most coastal areas experience high tides?

Alex777 [14]

It doesn't matter where you are. "Tides" happen twice (hi - lo - hi - lo) in about 24 hours and 50 minutes. Anywhere.

If it doesn't do two complete cycles in 24 hours 50 minutes, then it's not tide.

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2 years ago

Gravitational energy can be negative?

AlladinOne [14]

answer:

yes

explanation:

At a separation of the surface of Earth (r=6400km) gravity wants pull the test mass closer and closer. ... So the work done by gravity is NEGATIVE. The gravitational potential energy is negative because us trying to do the opposite of what gravity wants needs positive energy.

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3 years ago

You are riding a bicycle up a gentle hill. It is fairly easy to increase your potential

Dahasolnce [82]

True or false: while riding a bicycle up a gentle hill, it fairly easy to increase your potential energy, but to increase your kinetic energy would ...

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2 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

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3 years ago