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kobusy [5.1K]
2 years ago
5

50 point!! Will mark brainliest!! (Please help)

Physics
2 answers:
wariber [46]2 years ago
5 0

I agree with the other answer

What is the variable?

~the price is the variable.

What happens to demand?

~It'll go down. Since the price of snow blowers will increase then the quantity demanded will go down.

jok3333 [9.3K]2 years ago
4 0

What is the variable?

~<em>the</em><em> </em><em>price</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>variable</em><em>.</em>

What happens to demand?

~It'll go down. Since the price of snow blowers will increase then the quantity demanded will go down.

Hope this helps- have a good day bro cya)

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If a car travels 400m in 20 seconds how fast is it going? 20 m/s
scoundrel [369]

Answer:

20m/s

Explanation:

Speed = distance / time

Speed = 400/20

Speed = 20

5 0
3 years ago
What flows in electricity?<br> A. Electrons<br> B. Protons<br> C. Neutrons
nordsb [41]

Answer:

A.

Explanation:

Electrons

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Which word describes a goal that is poorly written?
stiv31 [10]

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6 0
3 years ago
Read 2 more answers
What is the critical angle of light traveling from benzene (n=1.501) into air?
Taya2010 [7]
The critical angle formula should be: sin^-1(1/n) 

 where "n" is 1.501 into the air

<span>The critical angle of light travelling from benzene, happens because the larger angles of incidence from the inside of the benzene has experienced the total internal reflection. </span>


4 0
3 years ago
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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