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scoundrel [369]

3 years ago

8

A 2.35 kg ball is attached to a ceiling by a3.53 m long string. The height of the room is5.03 m .The acceleration of gravity is

9.8 m/s2.
What is the gravitational potential energyassociated with the ball relative to the ceiling?Answer in units of J.

1 answer:

Studentka2010 [4]3 years ago

3 0

Answer:-81.29 J

Explanation:

Given

mass of ball $m=2.35 kg$

Length of string $L=3.53 m$

height of Room $h=5.03 m$

Gravitational Potential Energy is given by

$P.E.=mgh$

where h=distance between datum and object

here Reference is ceiling

therefore $h=-3.53 m$

Potential Energy of ball w.r.t ceiling

$P.E.=2.35\times 9.8\times (-3.53)=-81.29 J$

i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling

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A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.

nikdorinn [45]

Answer:

a

$\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

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From the question we are told that

The current is $I = 0.106 \ A$

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The separation between the plate is $d = 4.0 mm = 0.004 \ m$

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differentiating both sides with respect to t is

$\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}$

=> $\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I$

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$\epsilon _o = 8.85*10^{-12} C/(V \cdot m)$

=> $\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}$

=> $\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

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4 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a

ValentinkaMS [17]

Answer:

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to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

$E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m$

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σ=ξE

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c.

from the expression for the capacitance

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$Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC$

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4 years ago