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3 years ago

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Click to review the online content. Then answer the question(s) below, using complete sentences. Scroll down to view additional

questions. "The Histogram" Describe the function of a histogram and list three types of information that histograms convey. (Site 1)

2 answers:

Igoryamba3 years ago

6 0

Answer:

Rococo art relates to an artistic style especially of the 18th century characterized by fanciful curved asymmetrical forms and elaborate ornamentation.

Explanation:

yes

denpristay [2]3 years ago

6 0

Answer:

taking advantage on how old this is im sorry DFGHJKL:

Explanation:

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A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

<h3>a)</h3>

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

<h3>b)</h3>

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

<h3>c)</h3>

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

<h3>d)</h3>

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

4 years ago

In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 609 nm.

Ray Of Light [21]

Answer:

$\Delta x = 3.65 \mu m$

Explanation:

As we know that the sixth order maximum will have path difference given as

$\Delta x = N\lambda$

here we know that

N = order of maximum

$\lambda = 609 nm$

now we have

$N = 6$

so we know that

$\Delta x = 6(609 nm)$

$\Delta x = 3.65 \mu m$

6 0

3 years ago

Y=-3x +4 find the domain Nd range

aksik [14]

Is 3678.555667775 correct

3 0

2 years ago

How long will a bus take to travel 150 km at an average speed of 40 km h?

yan [13]

225 mins long will a bus take to travel 150 km at an average speed of 40 km h

Since the average speed is 40km/h it means

40 km can be travelled by it in 1 hour

1 km can be travelled by it in 1/40 hour

So 150km can be travelled by it in 150/40 hours or 225 mins.

By formula time=distance/speed=150/40=3.75hours

Speed = Distance/Time – This tells us how slow or fast an object moves. It describes the distance travelled divided by the time taken to cover the distance. Time = Distance / Speed, as the speed increases the time taken will decrease and vice versa.

To know more about average speed click here

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3 0

1 year ago

A student is studying a type of wave. Which question will help the student determine if it is an electromagnetic wave?

eimsori [14]

I believe D is the correct answer

5 0

3 years ago

Read 2 more answers