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Softa [21]
3 years ago
13

What is produced as the result of unequal warming of the earth's surface?

Physics
1 answer:
Soloha48 [4]3 years ago
7 0
Unusual precipitation patterns 
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Which forces tend to slow down an object
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The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop. 

Hope this helps!
8 0
4 years ago
The gasoline in a car does 40,000 J of work on a car and generates a constant force of 20 N. How far did the car go?
AnnyKZ [126]

L=F•d=>d=L/F=40,000/20=2,000 m

7 0
3 years ago
With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be
Delvig [45]

Answer:

The resistance must be 6.67\Omega

Solution:

Resistance, R_{1} = 20\Omega

Resistance, R_{2} = 10\Omega

For the current to be the same when the switch is open or closed, the resistances must be connected in parallel as current is distributed in parallel with the same voltage across the circuit:

Thus in parallel:

\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}

R_{eq} = 6.67\ \Omega

4 0
3 years ago
Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m
Kisachek [45]
<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>
8 0
3 years ago
what is the name of the tool that allows you to copy two or more shapes into a single part? when working with 3D
Anna [14]

Answer:

Modeling tool or Align tool. it depends what type of sandbox platform you use

Explanation:

1

8 0
3 years ago
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