All categories

2 years ago

How many earth years does it take mercury to go around the sun?

Physics

2 answers:

Bad White [126]2 years ago

5 0

<h2>Answer:</h2><h3>0.24 Earth years</h3><h3 /><h2>Explanation:</h2>

However, it's proximity to the Sun means that its average orbital velocity is a speedy 47.362 kilometers a second or 29.429 miles per second – approximately 170,500 km/h; 105,945 mph. At this rate, it takes Mercury 87.969 days, or the equivalent of 0.24 Earth years, to complete a single orbit of the Sun.

sertanlavr [38]2 years ago

5 0

88 days to orbit length of day 58d 15h 30 m

You might be interested in

PLEASE HELP ASAP! THANKS!<br><br> Why is it a rare occasion to see a solar eclipse from your home?

Y_Kistochka [10]

Answer:

It is very rare to see a solar eclipse from your home, because the Earth, Sun, and the moon need to align just right. Not everyone in the world can view a solar eclipse, only some area can. A solar eclipse is where the moon blocks out the sun. If you think about it: Let's say you live in Florida, U.S.A. You may see the moon coming in front of the sun, but if you lived in California or sumthin', the moon and the sun wouldn't be aligned to form a solar eclipse. It all depends on location... so it is rare to see one.

5 0

2 years ago

What is the definition of Cubit

disa [49]

Answer:

See below.

Explanation:

Cubit is a unit of length based on the length of the forearm from the elbow to the tip of the middle finger and usually equal to about 18 inches (46 centimeters).

It is an ancient unit of length used in ancient Egypt and is also known as "ancient Egyptian royal cubit."

Other similar units of measurements are displayed in the image below. Thanks!

7 0

2 years ago

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m × 0.55 m. The magnetic field has

Pani-rosa [81]

Answer:

Part a)

$EMF = 0.38 V$

Part b)

$\frac{dA}{dt} = 0.43 m^2/s$

Explanation:

Part a)

Initial value of magnetic flux is given as

$\phi_1 = BAcos\theta$

$\phi_1 = (2.1)(0.35 \times 0.55) cos65$

so we have

$\phi_1 = 0.17 Wb$

Final flux through the loop is given as

$\phi_2 = 0$

now EMF is given as

$EMF = \frac{\phi_1 - \phi_2}{\Delta t}$

$EMF = \frac{0.17 - 0}{0.45}$

$EMF = 0.38 V$

Part b)

If magnetic field is constant while Area is changing

So EMF is given as

$E = Bcos65 \times \frac{dA}{dt}$

$0.38 = 2.1 cos65(\frac{dA}{dt})$

$\frac{dA}{dt} = 0.43 m^2/s$

5 0

2 years ago

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The

lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s

Explanation:

1) We can find the speed of the object by conservation of energy:

$E_{i} = E_{f}$

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s$

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s$

3) When the object is at the equilibrium position, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s$

I hope it helps you!

8 0

2 years ago

Why is metal often used for cooking pans? Metals

Yanka [14]

Explain

Metals are good conductors of heat, which is important for cooking (both heating and cooling occurs rapidly

3 0

2 years ago