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3 years ago

The map below shows major ocean currents in the North Atlantic and North Pacific Oceans. In general, currents flowing toward the

Physics

2 answers:

Lina20 [59]3 years ago

4 0

Answer:

Explanation:

9966 [12]3 years ago

4 0

Answer:

The United States West Coast. Hope this helps :)

Explanation:

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A reaction in a solution would proceed slowest when -

Zinaida [17]

the answer is A *but I am not sure*

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3 years ago

The Celsius temperature of –273° C is termed “absolute zero” and is the initial value on the metric unit of temperature, the Kel

Kazeer [188]

Answer:

273 Kelvin

Explanation:

If -273 Celsius is 0 Kelvin, then 273 Kelvin will be 0 Celsius.

7 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = $\frac{k\ q_1\ q_2}{r^2}$ .....................1

and here k is 9 × $10^9$ N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = $\frac{k\ q^2}{d^2}$

and force between charges at A to C, at 2d distance

F1 = $\frac{k\ q^2}{(2d)^2}$ = $\frac{k\ q^2}{4d^2}$

force between charges at A to D, 3d distance

F1 = $\frac{k\ q^2}{(3d)^2}$ = $\frac{k\ q^2}{9d^2}$

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a) = -F1 - F2 + F3 ....................2

put here value

F(a) = $-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

solve it

F(a) = $\frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

F(a) = $-\frac{41}{36}\ F1$ = 1.13 F1

and

Charge b It receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2 ....................3

F(b) = $\frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}$

F(b) = $\frac{1}{4} \ F1$ = 0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1 ....................4

F(c) = $\frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}$

F(c) = $\frac{9}{4} \ F1$ = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1 ....................5

F(d) = $-\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}$

F(d) = $-\frac{49}{36} \ F1$ = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = $\frac{2.25}{0.25}$

ratio = 9 : 1

5 0

3 years ago

Think about the electricity sent to your home from a power plant. How does the voltage of the electricity that leaves the plant

tatiyna

C. High voltage to low voltage

6 0

3 years ago

PLEASEEEEE HELP ME !!!

Elza [17]

Answer:

maybe its heat sorry if it's wrong

because if friction is not in the problem so we are making heat or thermal energy

4 0

2 years ago