All categories

3 years ago

How are important properties of Mercury, Venus, and Mars different from important properties of Earth?

1 answer:

3 0

Hi there,

There is one thing that is the most difference about other planets. Earth is a perfect condition for humans, animals, and other living things. Venus and Mars are not quite good for that. Venus has a very HOT surface, which makes it nearly impossible to be on. Mars on the other hand is very COLD, colder than earth because it is farther from the sun.

You might be interested in

Define uniform and non uniform

cricket20 [7]

Answer:

When a body moves along a straight line with uniform speed or steady speed is called Uniform motion. When a body moves along a straight line but with variable or change in speed is called non-uniform motion.Hope this answer helps.

7 0

3 years ago

Read 2 more answers

What two factors are a part of thermohaline circulation

Margarita [4]

Answer:

These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation.

Explanation:

7 0

3 years ago

A person walks first at a constant speed of 5.50 m/s along a straight line from point A to point B and then back along the line

Sav [38]

Answer:

4.25 m/s

Explanation:

They walked the first distance at 5.50 m/s, then the same distance at 3 m/s.

Since the distances are equal, the average speed is simply the average of 5.50 and 3.

(5.50 + 3) / 2 = 4.25

Her average speed over the entire trip is 4.25 m/s.

8 0

4 years ago

If the sun in 148 million kilometers from the earth, how many minutes will it take the light from the sun to reach the earth?

Natasha2012 [34]

Answer:

about 8 mins and 20 secs

Explanation:

8 0

3 years ago

Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de G

MakcuM [25]

Answer:

The magnitude of the electric field is 1.124 X 10⁷ N/C

Explanation:

Magnitude of electric field is given as;

$E = \frac{kq}{r^2} , N/C$

where;

E is the magnitude of the electric field, N/C

q is the point charge, C

k is coulomb's constant, Nm²/C²

r is the distance of the point charge, m

Given;

q = 5mC = 5×10⁻³ C

r = 2m

k = 8.99 × 10⁹ Nm²/C²

Substitute these values and solve for magnitude of electric field E

$E = \frac{kq}{r^2} = \frac{(8.99 X10^9)(5X10^{-3})}{2^2} = 1.124 X10^7 N/C$

Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C

7 0

3 years ago