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2 years ago

To date, Saturn is known to have 13 rings made from which of the following?

Physics

2 answers:

mylen [45]2 years ago

7 0

Answer:

it is believed to be shattered moons, asteroids, and comets, so C

Explanation:

ankoles [38]2 years ago

5 0

I think the answer is Dust. Moons and stars definitely don't seem likely and dark particles, I am not even sure what those are. But I have seen rings on other planets before. Hope this helps. :)

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The study of ____ is called kinematics

weqwewe [10]

Motaion would be it have a good day

7 0

2 years ago

Read 2 more answers

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

3 0

2 years ago

A linear accelerator uses alternating electric fields to accelerate electrons to close to the speed of light. A small number of

kobusy [5.1K]

Answer:

8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.

Explanation:

The total charge accumulated in 1.8 hours will be:

Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)

Total Charge = - 12960 nC = - 12.96 x 10^(-6) C

Since, the charge on one electron is e = - 1.6 x 10^(-19) C

Therefore, no. of electrons will be:

No. of electrons = Total Charge/Charge on one electron

No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]

<u>No. of electrons = 8.1 x 10^13 electrons</u>

6 0

2 years ago

IN WHAT SEQUENCE DOES MAGMA MOVE THROUGH A VOLCANO?

Natalka [10]

Well the volcanos come from the lower crust and moves up with changes the climate

6 0

3 years ago

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Fed [463]

<h2>Option A is the correct answer.</h2>

Explanation:

Acceleration due to gravity

$g=\frac{GM}{r^2}$

G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

Let mass of earth be M and radius of earth be r.

We have

$g=\frac{GM}{r^2}$

Now

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Mass of hypothetical planet, M' = M/2

Radius of hypothetical planet, r' = 2r

Substituting

$g'=\frac{GM'}{r'^2}\\\\g'=\frac{G\times \frac{M}{2}}{(2r)^2}\\\\g'=\frac{\frac{GM}{r^2}}{8}\\\\g'=\frac{g}{8}$

Option A is the correct answer.

6 0

3 years ago