All categories

3 years ago

Help me with this???

Physics

1 answer:

Vikentia [17]3 years ago

8 0

Yo sup??

Average velocity=total distance covered/total time taken

total distance covered=4 + 8=12 miles

total time taken=6 hours

Therefore

average velocity=12/6

=2 miles/hour

Hope this helps

You might be interested in

The process by which the nuclei of radio active atoms are split into 2 or more smaller nuclei is called what

Andreas93 [3]

Are there multiple choice?

3 0

3 years ago

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:

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 $v$ :

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

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.

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

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

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

Solve the equation for $R$ :

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

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

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

3 years ago

particle moves over three quarters of a circle of radius R what is the magnitude of its displacement

fomenos

Answer:

is equal to threee idont know

Explanation:

hindi ko alam ang explanation kwa kwek kwen hahahahhahahahaha

sanay akoy pag padendyahan niyo nay. hahahahhahahaha maawa kayo saakin mga aking mahal hahahhahahahahhaha hahah en then go char hahahhaa

6 0

3 years ago

What is the average speed between 5 seconds and 15 seconds?

hichkok12 [17]

Answer:

10 seconds

Explanation:

We can calculate the average speed by finding the total distance traveled divided by the elapsed time: Average speed=–s=Total distance Elapsed time

3 0

3 years ago

Read 2 more answers

A metal, conducting cylinder of length 0.25 meters contains a total charge of 0.5 coulombs. Use Gauss' Law to find the electric

aliina [53]

Answer:

Explanation:

Let assume that cylinder has a radius of 0.01 meters. The Gauss Law has the following form:

$\frac{Q_{enc}}{\epsilon_{o}}= E\cdot A$

The electric field considering the cylinder configuration that encloses the conducting cylinder is:

$E = \frac{Q_{enc}}{\epsilon_{o}\cdot (2\cdot \pi \cdot R \cdot L)}$

Where $\epsilon_{o}$ is the absolute permittivity.

$E=\frac{0.5\,C}{\left(8.854\times 10^{-12}\,\frac{C^{2}}{N\cdot m^{2}} \right)\cdot (2\cdot \pi)\cdot (0.011\,m)\cdot (0.25\,m)}$

$E = 3.268\times 10^{12}\,\frac{N}{C}$

4 0

3 years ago