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2 years ago

What is my displacement if i walked 200 meters west then turned around and walked 150 meters east

1 answer:

6 0

You're walking in one direction, and then the exact opposite of that direction, so you simply have to subtract the two distances.
200-150=50
You're 50 meters west of where you originally started.
You're west because 200 meters west is greater than 150 meters east. If the distance walked east was greater than the distance walked west, you would've been east of your starting position.

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Which waves require a material medium for transmission? a.) light waves b.) radio waves c.) sound waves d.) microwaves

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Answer:

mechanical waves require a medium

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2 years ago

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Why give the ecuación

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2 years ago

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$

Now let´s plug in the values we know:

$F = 1 lb$ mutual gravitational attraction force
$G = 6.67(10)^{-11}$ gravitational constant
$ρ_{iron} =491.5 \frac{lb}{ft^{3} }$

$1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}$

Solve for r and multiply by 2 because 2r = diameter

$d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }$

Result is d = 21.81 Feet

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2 years ago

A proton traveling to the right enters a region of uniform magnetic field that points into the page.When the proton enters this

neonofarm [45]

Answer:

Explanation:

We can answer this problem by using Fleming's Left Hand Rule. By doing so, we have to place:

- The index finger of the left hand in the direction of the magnetic field

- The middle finger of the left hand in the direction of the particle's velocity

- The thumb will give the direction of the force, and therefore the deflection of the proton

In this problem, we have:

- Magnetic field direction: into the page --> index finger

- Proton's velocity: to the right --> middle finger

By doing so, we observe that the thumb points towards the top of the page: therefore, the correct answer is

(C) deflected toward the top of the page.

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2 years ago

If the swimmer could cross a 14 km channel maintaining the same average velocity as for the first 50 m in the pool, how long wou

alukav5142 [94]

Complete question:

If the swimmer could cross a 14 km channel maintaining the same average velocity as for the first 50 m in the pool, how long would it take?

For the first 50m in the pool, the average velocity was 2.08 m/s

Answer:

It would take for the swimmer approximately 1.87 hours.

Explanation:

If the swimmer maintains the average velocity on the channel, we should find and approximate value of the time it takes to cross the channel with the Galileo’s kinematic equation:

$x=vt$

With x the displacement, v the average velocity and t the time, solving for t:

$t=\frac{x}{v}=\frac{14000m}{2.08\frac{m}{s}}$

$t=6730.7 s= 1.87 h$

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3 years ago