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s344n2d4d5 [400]
3 years ago
7

The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist

and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF.A) Estimate the charge that passed through Musschenbroek's body.
B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?
Physics
1 answer:
skelet666 [1.2K]3 years ago
8 0

Answer:

a) q = 4.47 10⁻⁵ C

b)     ΔV = 4.47 10⁴ V

Explanation:

A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor

           U = Q² / 2C

         Q = √ (2UC)

let's reduce the magnitudes to the SI system

   c = 1 nF = 1 10⁻⁹ F

let's calculate

         q = √ (2 1 10⁻⁹-9)

         q = 0.447 10⁻⁴ C

         q = 4.47 10⁻⁵ C

b) for the potential difference we use

             C = Q / ΔV

            ΔV = Q / C

            ΔV = 4.47 10⁻⁵ / 1 10⁻⁹

            ΔV = 4.47 10⁴ V

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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 2.3 cm from the axis of rotation. (a) Calcul
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Answer:

a) a_{r} = 0.275\,\frac{m}{s^{2}}, b) \mu_{s} = 0.028, c) \mu_{s} = 0.036

Explanation:

a) The linear acceleration of the watermelon seed is:

a_{r} = \omega^{2}\cdot r

a_{r} = \left[\left(33\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\right]^{2}\cdot (0.023\,m)

a_{r} = 0.275\,\frac{m}{s^{2}}

b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:

\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a

a = \mu_{s}\cdot g

\mu_{s} = \frac{a}{g}

\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }

\mu_{s} = 0.028

c) Angular acceleration experimented by the turntable is:

\alpha = \frac{\omega-\omega_{o}}{\Delta t}

\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}

\alpha = 9.6\,\frac{rad}{s^{2}}

The tangential acceleration experimented by the watermelon seed is:

a_{t} = \left(9.6\,\frac{rad}{s^{2}} \right)\cdot (0.023\,m)

a_{t} = 0.221\,\frac{m}{s^{2}}

The linear acceleration experimented by the watermelon seed is:

a = \sqrt{a_{t}^{2}+a_{r}^{2}}

a = \sqrt{\left(0.221\,\frac{m}{s^{2}} \right)^{2}+\left(0.275\,\frac{m}{s^{2}} \right)^{2}}

a = 0.353\,\frac{m}{s^{2}}

The minimum coefficient of static friction is:

\mu_{s} = \frac{0.353\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }

\mu_{s} = 0.036

4 0
3 years ago
(a) A force F = (3xî + 4yĵ), where F is in newtons and x and y are in meters, acts on an object as the object moves in the x-dir
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Answer:

W = 53.6648 J

Explanation:

W = ∫ F * dr

F = < 4*x i , 3*y j > Newtons

dr = < dx, 0 >

Take the dot product:

F * dr = 4* x * dx

Now replacing numeric

W = ∫ 4 * x * dx , ║ x = ( 0 m ⇒ 5.18 m )

W = ¹ / ₂ * (4 N/m) * x ²  

W = (2 N/m) * (5.18 m)²

W = 53.6648 J

8 0
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One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

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Since the rope is unstretchable, the total length will always be 34m.

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L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
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