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2 years ago

If you ran 15 km/h for 20 min, how much distance would you cover?

1 answer:

4 0

Since we have 15 kilometers per hour, and we're looking for 20 minutes, let's set up proportions.
20/60 minutes = x/15
20/60 = 1/3, so let's leave that simplified.
1/3 = x/15
Look at the denominators, 3 to 15 is a factor of 5, so multiply the numerator by 5.
1 • 5 = 5, so you will cover 5 kilometers in 20 minutes.

I hope this helps!

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

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3 years ago

Which of the following solutions should be chosen to help restore an ecosystem naturally? A. reintroducing an animal to the ecos

jolli1 [7]

Answer:

A. reintroducing an animal to the ecosystem

Explanation:

As generally, all know that for restoring an ecosystem naturally, it requires reintroduction of an animal to the ecosystem. As though it helps in reimposing the ecosystem back, and also helps to improve our ecosystem in natural surroundings, natural terrain, and population density. Basically reintroducing an animal is also required for the balancing of the ecosystem. As everything requires a properly balanced nature.

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2 years ago

Which statement describes a scientific law

Olegator [25]

The Earth is a sphere is a statement to describe a scientific law

8 0

2 years ago

When you whirl a can at the end of a string in a circular path, what is the direction of the force that acts on the can

andreyandreev [35.5K]

Answer:

Toward the centre of the circular path

Explanation:

The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).

In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by

$F=m\frac{v^2}{r}$

where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.

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2 years ago

Select the correct answer.

scZoUnD [109]

Answer:

B. It is directly proportional to the source charge.

Explanation:

Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.

This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.

This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.

Mathematically, Gauss's law is given by this formula;

ϕ = (Q/ϵ0)

Where;

ϕ is the electric flux.

Q represents the total charge in an enclosed surface.

ε0 is the electric constant.

Hence, the statement which is true of the electric field at a distance from the source charge is that it is directly proportional to the source charge.

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2 years ago