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4 years ago

What is the relationship between Mercury’s rotational period and orbital period?

Physics

1 answer:

Yuri [45]4 years ago

8 0

Answer and Explanation:

The speed of the mercury is exactly 2/3 of the orbital speed, We know that mercury rotate ones in 58.647 days and the orbital speed is 87.97 days so from here the mercury speed is 2/3 of the orbital speed.

The speed of mercury $=\frac{2}{3}$ orbital speed this is the relation between the mercury speed and orbital speed

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What is the net force on an object that has a force pushing downward at 25N and a forcing pushing upward at 10N?

shutvik [7]

Answer:

15N downwards

Explanation:

Net force is the sum of all forces acting on an object.

Taking upwards as positive and downwards as negative,

Net Force = 10N + (-25N)

= -15N

Hence <u>Net Force is 15N downwards</u>.

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4 years ago

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Q.1 Newton's second law gives the measure of--------------. 1)Acceleration 2) Force 3) Momentum 4)Angular momentum

posledela

Answer:

1. Acceleration

Explanation:

Newtons Second law gives the measure of acceleration

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4 years ago

WHAT DOES THE INCREASE IN TEMPERATURE INDICATE

KATRIN_1 [288]

Increase in temperature means:
- The substance is getting hotter
- It's internal energy is rising

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3 years ago

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SIL. TUILE I JUMU<br>Write down your own examples for each of the different effects of forces.

Harrizon [31]

Answer:

magnetic force

Explanation:

when some one gets charged up by maybe static eletricity the person will sustain the energy and the energy will make a force field around the person which can and will attract paper and he will be attractive

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3 years ago

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A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .

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3 years ago