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cricket20 [7]
3 years ago
14

Eight joules of work is required to stretch a spring 0.5 meter from its natural length. Find the work required to stretch the sp

ring an additional 0.40 meter.
Physics
1 answer:
Alexxandr [17]3 years ago
8 0

Additional Work done to stretch the spring 0.4m is 6.4J

<u>Explanation:</u>

<u></u>

Given -

Work done to stretch a spring, W1 = 8J

Distance , d = 0.5m

Distance2, x = 0.4m

Work done 2, W2 = ?

Using Hooke's law,

Work done = k x

where k is spring constant

For distance 0.5m :

8 J = k X 0.5m

k = 8/0.5 = 16 J/m

For distance 0.4m:

W2 = 16 X 0.4 J

W2 = 6.4J

Additional Work done to stretch the spring 0.4m is 6.4J

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An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

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