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2 years ago

In outer space, a piece of rock continues moving at the same velocity for

1 answer:

7 0

The absence of external force in the outer space, allows the piece of rock to continue moving at the same velocity for thousands of years.

<h3>Absence of external force on the outer space</h3>

The outer space is almost an absolute vacuum, because it's nearly empty. There is no matter such as air in the outer space that will provide an external force needed to change the velocity of the piece of rock.

From Newton's first law of motion, an object in a state of rest or uniform motion in a straight line, will continue in that state unless it is acted upon by an external force.

Thus, the absence of external force in the outer space, allows the piece of rock to continue moving at the same velocity for thousands of years.

Learn more about outer space here: brainly.com/question/24701339

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HELP

Ivanshal [37]

Answer:

The velocity is 60 km/hr.

Explanation:

<h3><u>Given:</u></h3>

Displacement (d) = 480 km = 48000 m

Time (t) = 8 Hours = 480 minute

Velocity (v) = ?

Now,

Velocity = Displacement ÷ Time

v = d/t

v = 480/8

v = 60 km/hr

Thus, The velocity is 60 km/hr.

<u>-TheUnknownScientist 72</u>

5 0

3 years ago

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.

Thus,

r = $\frac{mv}{qB}$ ÷ sinΘ

But, sinΘ = sin $90^{0}$ = 1.

So that;

r = $\frac{mv}{qB}$

= (9.11 × $10^{-31}$ × 121) ÷ (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ )

= 1.10231 × $10^{-28}$ ÷ 2.608 × $10^{-24}$

= 4.2266 × $10^{-5}$

= 4.23 × $10^{-5}$ m

The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

B. The frequency 'f' of the motion is called cyclotron frequency;

f = $\frac{qB}{2\pi m}$

= (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ ) ÷ (2 × $\frac{22}{7}$ × 9.11 × $10^{-31}$ )

= 2.608 × $10^{-24}$ ÷ 5.7263 × $10^{-30}$

= 455442.4323

f = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0

4 years ago

Read 2 more answers

Planet X is a terrestrial planet in our solar system. It has 21% oxygen in its atmosphere. Humans can walk on this planet withou

Ulleksa [173]

Answer:

a planet the is human habitable or just plain out earth

Explanation:

8 0

3 years ago

Read 2 more answers

WILL GIVE 5 STARS!!! HELP ASAP!!!

DedPeter [7]

Most fish fossils are: <u>D. from sharks, eels, and stingrays</u>.

<u>Explanation</u>:

A fossil is a preserved part or trace of a living being in the geological age. Animals, fish, bones and shells are some of the fossils. Fish fossils are the dead fish that are found under the river bed or ocean.

The dead animals in the water bodies are buried under the mud for long period of time. The soft tissues of the animals are decomposed leaving the dry bones.

There are five different types of fossils. They are

i) Body fossils

ii) Molds and casts

iii) Petrification fossils

iv) Footprints and trackways

v) Coprolites

One of the common fossils is shark tooth. The tooth of the sharks is shed continuously. The fish fossils are found at the bottom of the ocean, lake and pond deposits.

8 0

4 years ago

The intensity of sunlight at the Earth is about 1367 W/m^2. How many times further away would the Sun have to be from the Earth

grandymaker [24]

The intensity of sunlight on the Earth is about 1367 W/m^2. The earth has to be 131 the distance farther away

This is further explained below.

<h3>What is the distance?</h3>

Generally, bet the power emitted by the sun is P. so. at the distance of earth intensely where $r_E$ is the distance of the earth from the sun. were $I_1=1362 \frac{w}{m^2}$

Therefore

p=4 $\pi 1367 r_E^2$

let at distance $\gamma r_E$$ is a positive number) the intensity is the same as the intensity at 10m away from a 100w bulb. So.

In conclusion,

$\begin{aligned}&\frac{P}{4 \pi\left(\gamma r_{\varepsilon}\right)^2}=\frac{100}{4 \pi 10^2}\\\\&\text { or, }\left(\gamma r_E\right)^2=P=4 \pi 1367 r_E^2\\\\&\therefore \gamma=\sqrt{4 \pi \times 1367}\\\\&=131.06\\\\ }\end{aligned}$