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3 years ago

How long does it take mercury to orbit the sun?

2 answers:

5 0

Mercury is the smallest planet of the eight planets and is made up of rock and metal. One year on Mercury is only 88 Earth days. That's how long it takes for Mercury to orbit the sun. Mercury rotates very slowly, so its days are very long. A day on Mercury is 59 Earth days.

Oliga [24]3 years ago

3 0

That is called the "Revolution Period of a planet" and one year on mercury equals to 88 days as compared to Earth

In short, Your Answer would be 88

Hope this helps!

You might be interested in

An objects speed is the distance it travels____the amount of times it takes?

beks73 [17]

Multiplied by; speed = distance x time

8 0

4 years ago

Please help me people

saveliy_v [14]

Answer :

(1) The density of asphalt is, $1200kg/m^3$

(2) (a) Length, width and thickness of sheet in meter is, 0.35 m, 1.1 m and 0.015 m respectively.

(b) The volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

Explanation :

Part 1 :

As we are given:

Mass of block = 90 kg

Volume of block = $0.075m^3$

Formula used :

$\text{Density of block}=\frac{\text{Mass of block}}{\text{Volume of block}}$

Now put all the given values in this formula, we get:

$\text{Density of block}=\frac{90kg}{0.075m^3}=1200kg/m^3$

Thus, the density of asphalt is, $1200kg/m^3$

Part 2(a) :

As we are given that:

Length of aluminium sheet = 35 cm

Width of aluminium sheet = 11 dm

Thickness of aluminium sheet = 15 mm

Now we have to convert these dimensions into meters.

Conversions used:

1 cm = 0.01 m

1 dm = 0.1 m

1 mm = 0.001 m

Length of aluminium sheet = 35 cm = 35 × 0.01 = 0.35 m

Width of aluminium sheet = 11 dm = 11 × 0.1 = 1.1 m

Thickness of aluminium sheet = 15 mm = 15 × 0.001 = 0.015 m

Part 2(b) :

First we have to calculate the volume of aluminium sheet.

Volume of aluminum sheet (cuboid) = Length × Width × Thickness

Volume of aluminum sheet (cuboid) = 0.35 m × 1.1 m × 0.015 m

Volume of aluminum sheet (cuboid) = 0.005775 m³

Now we have to calculate the mass of aluminium sheet.

$\text{Density of aluminium}=\frac{\text{Mass of aluminium}}{\text{Volume of aluminium}}$

$2700kg/m^3=\frac{\text{Mass of aluminium}}{0.005775m^3}$

$\text{Mass of aluminium}=15.59kg$

Thus, the volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

7 0

3 years ago

Read 2 more answers

g Show that the first derivative of the magnitude of the net magnetic field of the coils (dB/dx) vanishes at the mid- point P re

never [62]

Answer:

Hi Carter,

The complete answer along with the explanation is shown below.

I hope it will clear your query

Pls rate me brainliest bro

Explanation:

The magnitude of the magnetic field on the axis of a circular loop, a distance z from the loop center, is given by Eq.:

B = NμοiR² / 2(R²+Z²)³÷²

where

R is the radius of the loop

N is the number of turns

i is the current.

Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense, and the fields they produce are in the same direction in the region between them. We place the origin at the center of the left-hand loop and let x be the coordinate of a point on the axis between the loops. To calculate the field of the left-hand loop, we set z = x in the equation above. The chosen point on the axis is a distance s – x from the center of the right-hand loop. To calculate the field it produces, we put z = s – x in the equation above. The total field at the point is therefore

B = NμοiR²/2 [1/ 2(R²+x²)³÷² + 1/ 2(R²+x²-2sx+s²)³÷²]

Its derivative with respect to x is

dB /dx= - NμοiR²/2 [3x/ (R²+x²)⁵÷² + 3(x-s)/(R²+x²-2sx+s²)⁵÷² ]

When this is evaluated for x = s/2 (the midpoint between the loops) the result is

dB /dx= - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷² - 3(s/2)/(R²+s²/4)⁵÷² ] =0

independent of the value of s.

8 0

3 years ago

Each proton-proton cycle generates 26.7 MeV of energy. If 9.9 Watts are generated via by the proton-proton cycle, how many billi

lina2011 [118]

Answer:

4.635 *10^12 Neutrinos

Explanation:

Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.

We proceed as follows;

In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.

From the question, we are given that

Power P = 9.9 watts = 9.9 J/s

Watts is same as J/s

The number of proton-proton cycles required to generate E energy is N = E / E '

Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question

Converting Mev to J, we have

= 26.7 x1.6 x10 -13 J

To get the number N which is the number of proton-proton cycle required, we have;

N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12

Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.

Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos

8 0

3 years ago

I just need an answer ASAP

nikdorinn [45]

Answer: c

Explanation: hope this helps :)

8 0

3 years ago

Read 2 more answers