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3 years ago

How long does it take for mars to rotate on its axis

Physics

2 answers:

Svetllana [295]3 years ago

8 0

Hello There!

It takes the planet Mars around 24 hours, 37 minutes, 23 seconds to rotate on its axis. This is around the same amount of time that it takes our planet to rotate once on its axis.

ratelena [41]3 years ago

3 0

Answer: 26.4 hours
Explanation: The axis of Mars is tilted at 25 degrees and 12 minutes relative to its orbital plane about the Sun. This produces seasons on the surface of Mars, similar to the seasons on Earth.

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The magnetic force on a straight wire 0.69 m long is 1.5 x 10-3 N. The current in the wire is 16.9 A. What is the magnitude of t

VikaD [51]

Answer:

Magnitude of magnetic field is 1.29 x 10⁻⁴ T

Explanation:

Given :

Current flowing through the wire, I = 16.9 A

Length of wire. L = 0.69 m

Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N

Consider B be the applied magnetic field.

The relation to determine the magnetic force experienced by current carrying wire is:

F = ILBsinθ

Here θ is the angle between magnetic field and current carrying wire.

According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:

F = ILB

$B=\frac{F}{IL}$

Substitute the suitable values in the above equation.

$B=\frac{1.5\times10^{-3} }{16.9\times0.69}$

B = 1.29 x 10⁻⁴ T

7 0

3 years ago

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

$a^3= \frac{T^2 GM}{4\pi^2}$

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

$R = 1.86303*10^{13}(1-0.997)$

$R= 5.58*10^{10}m$

7 0

3 years ago

• How does AC work?

Juli2301 [7.4K]

AC reverses the current in periods, and the current flows in 2 directions.
DC only flows in one direction and is constant.

AC is typically used to transfer power over long distances from a generator to your home, stepping it up after it's generated to send it over the distance, then stepping it back down so that it is usable in your home.

DC is used typically inside of devices, as it sends a certain amount of electricity through the wires.

Nicolas Tesla, was ripped off by Edison to replace his DC stations with AC instead. ^^

4 0

3 years ago

A bird on a long migration flies 63 kilometers per hour for 2900

Fed [463]

Answer:

Explanation:

2900 divided by 63 (2900/63)

- 46.03174603

Rounded to the nearest whole number

is 46.

THE ANSWER IS 46

5 0

3 years ago

A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, movin

padilas [110]

Answer:

To the right relative to the original frame.

Explanation:

In first reference frame S,

Spatial interval of the event, $\rm \Delta x=1000\ m-0\ m=1000\ m.$

Temporal interval of the event, $\rm \Delta t = 1\ \mu s=10^{-6}\ s.$

In the second reference frame S', the two flashes are simultaneous, which means that the temporal interval of the event in this frame is $\rm \Delta t'=0\ s.$

The speed of the frame S' with respect to frame S = v.

According to the Lorentz transformation,

$\rm \Delta t'=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right ).\\\\Since,\ \Delta t'=0,\\\therefore \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right )=0\\\Rightarrow \Delta t-\dfrac{v\Delta x}{c^2}=0\\\dfrac{v\Delta x}{c^2}=\Delta t\\v=\dfrac{c^2\Delta t}{\Delta x }.\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.$

And positive v means the velocity of the second frame S' is along the positive x-axis direction, i.e., to the right direction relative to the original frame S.

7 0

2 years ago