5. This break-dancer's speed is not changing as he spins on his head, but he is

I have the freedom of speech guaranteed by the 1st Amendment in the US Constitution; can I say whatever I want, whenever I want?

denis23 [38]

Answer:

No.

Explanation:

Even though it's called free speech, it really isn't free. You can't say certain things in public or on social media because people are too sensitive these days to handle the truth about something. You can, however, say what you want in a book or a protest with like-minded people because you really can't get in trouble for supporting whatever you want.

Hope this helps!

5 0

3 years ago

Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.

nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is $\frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2$

<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

Translational kinetic energy. Due to the linear movement of the body

$K_{tras} =\frac{1}{2} m v^2$

Rotational kinetic energy. Due to the rotational movement of the body.

$K_{rot} = \frac{1}{2} I w^2$

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

$K_{earth} = K_{tras} + K_{rot}$

Linear and rotational speed are related.

v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

I = $\frac{2}{5 } m r^2$

Let's subatitute.

$K_{earth} = \frac{1}{2} \ m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}$

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

$w = \frac{2 \pi}{T}$

Let's substitute.

$K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2 \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2$

$K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)$

Data for Earth are tabulated:

Mass m = 5.98 1024 kg

Radius r = 6.37 10⁶ m

Radius orbits tras = 1.496 10¹¹ m

Rotation period $T_{rot}$ = 24 h ( $\frac{3600s}{1h}$ ) = 8.64 10⁴s

Translation period $T_{tras}$ = 365 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 3.15 10⁷ s

Let's calculate.

$K_{earth} = 4 \pi^2 5.98 \ 10^{24} ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2 \ + \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )$

$K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)$

$K_{earth}= 2.66 \ 10^{33} J$

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of $T_{Sum}$ = 27 days.

The kinetic energy of the sun is;

$K_{sum} = K_{rot} = \frac{1}{2} I w^2$

$K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2$

$K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2$

The tabulated data for the sun are:

Mass m = 1,991 1030 kg.

Radius R = 6.96 10⁸ m

Period T = 27 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 2.33 10⁶ s

Let's calculate.

$K_{sum} = 1.40 \ 10^{36} J$

The relationship of the kinetic energy of the sun and the Earth is:

$\frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}$

$\frac{K_{sum}}{K_{earth}} = 5.3 \ 10^2$

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is: $\frac{K_{Sum}}{K_{Earth}} = 5 \ 10^2$

Learn more about kinetic energy here: brainly.com/question/25959744

5 0

3 years ago

A moving object has a kinetic energy of 150 j and a momentum with a magnitude of 30.0 kg•m/s. determine the mass and speed of th

SOVA2 [1]

To determine the answer to this item, we use two (2) equations.

Equation for kinetic energy:
KE = 0.5 mv²

Equation for momentum:
P = mv

From the second equation, we can deduced that,
m = P/v
Substituting the known values from the given above,

m = 30/v

Using this expression in the first equation,

KE = 0.5 mv²; 150 = 0.5(30/v)(v²)

The value of v from the equation is 10 m/s.

The mass is therefore calculated as such,
m = 30/v = 30/10 = 3 kg

Hence, the answers are,

<em> Mass = 3 kg</em>
<em> Velocity = 10 m/s</em>

4 0

3 years ago

At highway speeds, a particular car is capable of an acceleration of about 1.6 m/s?. At this rate,

kicyunya [14]

Given parameters:

Acceleration of the car = 1.6m/s

Initial speed = 80km/hr

Final speed = 110km/hr

Solution:

Time taken to achieve this speed = ?

Solution:

Acceleration is the rate of change of velocity with the time taken.

Mathematically;

a = $\frac{V - U}{T}$

where a is the acceleration

V is the final velocity

U is the initial velocity

T is the time taken

Now make the unknown time the subject of the expression;

aT = V - U

T = $\frac{V - U}{a}$

Convert the given acceleration to km/hr;

1.6m/s = 1.6 x $\frac{m}{s}$ x $\frac{1km}{1000m}$ x $\frac{3600s}{1hr}$ = 5.76km/hr

Input the parameters and solve;

T = $\frac{110 - 80}{5.76}$ = 5.2hrs

The time taken is 5.2hrs

5 0

4 years ago

What is the most effective way to prevent beach erosion?

sergiy2304 [10]

The most effective way to prevent beach erosion is to replant mangrove trees, seaweeds and seagrass. The strategic planting of vegetation can be used to help control erosion. The roots of these plants help to anchor the sand and ensure that it is not carried off in erosion. This is why many areas plant seagrass and build marshes along coasts to prevent erosion.

8 0

4 years ago