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Leona [35]
3 years ago
8

An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 N/m. Two small charged objects are attached

to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.022 m relative to its unstrained length. Determine (
a. the possible algebraic signs and (
b. the magnitude of the charges.
Physics
1 answer:
Kruka [31]3 years ago
8 0
The solution you should use is Hooke's law: F=-kx

It should have the same signs because they repel due to the stretch of the spring. 

a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be 
<span>F = kx 
270 N/m x 0.38 m = 102.6 N 
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force. 



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What is the speed of a car that travels 50m in 25min? *
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Tony drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took hours. When Tony drove hom
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Answer:

Question not completed, so I analysed the question first

Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?

Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

Let the speed to while going to the mountain be a miles/hour

Then, while going home it was b miles/hour faster than while going to the mountain.

Then, speed going home is (a+b)miles / hour

The formula for speed is given as

Speed=distance/time

The constant through out the journey is distance, the two journey has the same distance.

Then,

Distance =speed×time

For first journey going to the mountain

Distance = a×x=ax miles

For the second journey going home

Distance =y×(a+b)

Distance Mountain= distance home

ax=y(a+b)

Make a subject of the formula

ax=ya+yb

ax-ya=yb

a(x-y)=yb

a=yb/(x-y)

Therefore, distance from mountain is

Distance=speed ×time

Distance= a×x=ax

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So from the questions

x=6hours, y=4hours

Also, b=22miles/hour

Then,

a=yb/(x-y)

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a=44miles/hour

Then, the house distance from the mountain is

Distance=ax

Distance =44×6

Distance =264miles

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