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vaieri [72.5K]
2 years ago
12

Two identical vertical springs S1 and S2 have masses m1 = 400 g and m2 = 800 g attached to them. If m1 causes spring S1 to stret

ch by 4 cm, what is the ratio of the potential energy of S1 and S2? Use g = 10 m/s^2
Select one:

a. 1:2

b. 4:1

c. 1:4

d. 1:3

e. 2:1​
Physics
1 answer:
OlgaM077 [116]2 years ago
6 0

Answer:

potential energy = mgh

= 400÷1000 × 10× 4÷100

= 0.4 × 10 × 0.04

=4/10 ×10×4/100

= 4/10 × 4/10

=16/100

= 0.16 joules

m1 (400) stretches 4cm

m1 (100g) stretches 1cm

so, m2(800g) stretches 8 cm

potential energy of m2 = mgh

= 800/1000 ×10×8/100

= 0.8 × 0.8

=8/10 ×8/10

= 64/100

=0.64 joules

Ratio of s1 to s2

16/100 ÷ 64/100

= 1:4 ( answer)

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Four point charges, each of magnitude 2.38 µC, are placed at the corners of a square 75.2 cm on a side. If three of the charges
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Answer:

The Electric Force on Negative Charge is 2.968 N

Explanation:

charge on each corner, q = 2.38 micro coulomb

Side of square, a = 75.2 cm

Coulombic constant, K = 8.98755 x 10^9 Nm²/C²

sides of the square are A,B,C and D

and all sides of a square are equal so

AB = BC = CD = DA = 75.2 cm = 0.752 m

Diagonal, AC = BD = 1.414 x 0.752 = 1.06 m

Electric field at D due to charge at A

EA= Kq÷AB^2

= 8.98755×10^9 × 9.87×10^-6 ÷ 0.752^2

EA= 156863.82 N/C

Similarly Electric field at D due to charge C

EC=Kq÷CD^2

= 8.98755×10^9 ×9.87×10^-6 ÷ 0.752^2

EC= 156863.82 N/C

Electric field at D due to charge at BB

EB=Kq÷BD^2

EB=8.98755×10^9 × 9.87×10^-6 ÷ 1.06^2

EB=78949.01 N/C

Resolve the compoents

Ex = EA + EB cos 45

Ex = 156863.82 + 78949.01 x 0.707

Ex = 212689.2 N/C

Ey = EC + EB Sin 45

Ey = 156863.82 + 78949.01 x 0.707

Ey = 212689.2 N/C

The resultant electric field is

E = 1.414 x 212689.2 = 300787.95 N/C

the electric force on the negative charge is

F = q x E

F = 9.87 x 10^-6 x 300787.95

F = 2.968 N

7 0
2 years ago
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