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Novosadov [1.4K]
3 years ago
8

A toy gun uses a spring to project a 6.4 g soft rubber sphere horizontally. The spring constant is 9.0 N/m, the barrel of the gu

n is 12.4 cm long, and a constant frictional force of 0.028 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 5.0 cm for this launch? (Assume the projectile is in contact with the barrel for the full 12.4 cm.)
Physics
1 answer:
8_murik_8 [283]3 years ago
5 0

Answer:

1.584 m/s.

Explanation:

Using,

Ek= Ws-Wf...................... Equation 1

Where Wf = work done against friction, Ek = kinetic energy of the rubber, Ws = work done by the spring.

1/2mv² = 1/2ke²-F'd................ Equation 2

Where m = mass of the rubber, v = velocity of the rubber, k = force constant of the spring, e = compression, F' = force of friction, d = distance/ length of contact.

make v the subject of the equation

v = √[2(1/2ke²-F'd)/m].................... Equation 3

Given: k = 9 N/m. e = 5 cm = 0.05 m, F' = 0.028 N, d = 12.4 cm = 0.124 m, m = 6.4 g = 0.0064 kg

Substitute into equation 3

v = √[2(1/2×9×0.05²-0.026×0.124)/0.0064]

v = √(2[0.01125-0.003224]/0.0064)

v = √2.508125

v = 1.584 m/s.

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The diameter of the disks is 1.32 cm.

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Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C

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The capacitance of a parallel plate capacitor is given by,

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Rewrite the expression for A.

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the area A of the disks is given by,

A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }

Here, D is the diameter of the disk.

D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm

The diameter of each disc is found to be 1.32 cm.



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