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tino4ka555 [31]

3 years ago

12

What obstacles or challenges are encountered in magnetic levitation?

1 answer:

olasank [31]3 years ago

5 0

Mike Hawk Moe Lester Hugh Janice

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If a 126 gram marble rolls down a hill at an acceleration of 3.0m/s2, what would the force be on the marble?

adelina 88 [10]

Answer:

Explanation:

Net force would be the one causing the acceleration or

F = ma = 0.126(3.0) = 0.378 N

4 0

3 years ago

A 50-foot flagpole is at the entrance of a building that is 300 feet tall. If the length of the flagpole's shadow is 30 feet at

natka813 [3]

First you will want to sketch out both of the situations. It should be two sketches, one for the flagpole and one for the building.

To solve this, you will want to create a proportion.

Flagpole height/flagpole shadow=building height/building shadow

Therefore, it should look like this:
50/30= 300/x

Solve for x:
50x=9,000
X= 180 feet

YOUR ANSWER IS C.

7 0

3 years ago

Calculate the average speed of the moon around the earth. The moon has a period of revolution of 27.3 days and an average distan

klio [65]

<h2>The average speed of the moon around the earth is 1021.74 m/s</h2>

Explanation:

Radius of moon around earth, r = 3.84 × 10⁸ m

Circumference of orbit = 2πr = 2 x 3.14 x 3.84 × 10⁸

Circumference of orbit = 2.41 x 10⁹ m

Time taken, t = 27.3 days = 27.3 x 24 x 60 x 60 = 2358720 seconds

We have

Circumference of orbit = Speed of moon x Time taken

2.41 x 10⁹ = Speed of moon x 2358720

Speed of moon = 1021.74 m/s

The average speed of the moon around the earth is 1021.74 m/s

5 0

3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

Nina [5.8K]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

3 0

4 years ago

suppose a log's mass is 5 kg. After burning, the mass of the ash is 1 kg. explain what May have happened to the other 4 kg.

Katyanochek1 [597]

The other 4 kg may have left the scene in the form of
gases and smoke particles.

6 0

3 years ago