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Tju [1.3M]
3 years ago
6

What does Hooke's law say about the relationship of a spring to the force applied to it?

Physics
1 answer:
hoa [83]3 years ago
6 0

Answer:

Explanation:Hooke's law states that " if the elastic limit of an elastic material is not exceeded, the extension,e is directly proportional to the applied force or load.

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Two identical plastic cups contain the same amount of water at two different temperatures, as shown to the left. Both cups are p
Elza [17]

Answer:

the molecules will begin to move slowly and will turn to ice

Explanation:

hope this was good or not not sure if am right but yeah

6 0
3 years ago
Which situation best describes the act of reducing?
pishuonlain [190]
In my opinion, i think the first one is the best one since we're cutting down on paper wasted in mails when its better to do it online.

"paying bills online instead of sending paper through mail"
3 0
3 years ago
This type of reaction is most like breaking a rack of balls at the start of a game of pool:
Svet_ta [14]
The answer is b fission hope this helps

5 0
3 years ago
A friend of yours who is traveling the world reports that she recently experienced a day on which the surface wind blew from the
klasskru [66]

Answer:

At that time, she was in the southern hemisphere

Explanation:

She is witnessing the southeastern trade breeze because the cloud is passing from east to west and she is also witnessing a storm from the southeast.

Because of the Coriolis effect, which deviates the atmosphere in the left-hand side of the southern hemisphere and generates a southeastern wind, this wind did not blow directly from south to east.

4 0
3 years ago
A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =
Elis [28]

Answer:

F_{3h}=39065.298\times 10^9\ N attractive toward +x axis is the net horizontal force

F_v=80062.47\times 10^9 attractive toward +y axis is the net vertical force

Explanation:

Given:

  • charge at origin, Q_0=-3.35\times 10^{-6}\ C
  • magnitude of second charge, Q_2=2.05\times 10^{-6}\ C
  • magnitude of third charge, Q_3=5\times 10^{-6}\ C
  • position of second charge, (x_2,y_2)\equiv(0,4.35)\ cm
  • position of third charge, (x_3,y_3)\equiv(3.1,3.8)\ cm

<u>Now the distance between the charge at at origin and the second charge:</u>

d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}

d_2=\sqrt{(0-0)^2+(4.35-0)^2}

d_2=0.0435\ m

<u>Now the distance between the charge at at origin and the third charge:</u>

d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}

d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}

d_3=0.04904\ m

<u>Now the force due to second charge:</u>

F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}

F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}

F_2=32175.98\times 10^9\ N attractive towards +y

<u>Now the force due to third charge:</u>

F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}

F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}

F_3=61748.38\times 10^9\ N attractive

<u>Now the its horizontal component:</u>

F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9

F_{3h}=39065.298\times 10^9\ N attractive toward +x axis

<u>Now the its vertical component:</u>

F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9

F_{3v}=47886.49\times 10^9\ N upwards attractive

Now the net vertical force:

F_v=F_{3v}+F_2

F_v=47886.49\times 10^9+32175.98\times 10^9

F_v=80062.47\times 10^9

3 0
3 years ago
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