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GarryVolchara [31]
3 years ago
6

The clouds that occur at the highest altitude are usually

Physics
2 answers:
babunello [35]3 years ago
7 0
<span> cirrus hope this helps </span>
gogolik [260]3 years ago
5 0
The answer is A. Cirrus clouds occur at the highest altitude.
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Imagine a 10kg block moving with a velocity of 20m/s to the left.
lara31 [8.8K]
Kinetic energy , KE= [1/2]m*v^2

m = 10 kg
v=20m/s

KE = [1/][(10kg)(20m/s)^2 = [1/2](10kg)(400m^2/s^2) = 2000 joule

Answer: 2000 joule
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4 years ago
The description for a certain brand of house paint claims a coverage of 475 ft²/gal.
AlexFokin [52]

Answer:

(a) 11.66 square meters per liter

(b) 11657.8 per meters

(c) 0.00211 gal per square feet

Explanation:

(a) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 1gal/3.7854L = 11.66m^2/L

(b) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 264.17gal/1m^3 = 11657.8/m

(c) Inverse of 475ft^2/gal = 1/475ft^3/gal = 0.00211gal/ft^3

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3 years ago
What Does This Poem Mean?
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During what era did the French and Indian War occur?
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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
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