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3 years ago

Please can someone solve this for me i will really appreciate it

Physics

1 answer:

Blizzard [7]3 years ago

3 0

Answer:

• Average current = 25,000 Amperes

• Time = 0.80 seconds

${ \rm{quantity \: of \: charge = current \times time}} \\ { \rm{q = 25000 \times 0.80}} \\ { \boxed{ \mathfrak{answer : \: { \rm{quantity \: of \: charge = 20000 \: coulombs}} }}}$

${ \rm{number \: of \: electrons = \frac{current}{charge \: quantity} }} \\ \\ { \rm{n = \frac{25000}{20000} }} \\ \\ { \boxed{ \mathfrak{ \: answer : \: { \rm{n = 1.25 \: }}}}}$

${ \underline{ \mathfrak{ \green{Good \: grace}} \: \:⚜ \: \: { \blue{ \mathfrak{Good \: God}}}}}$

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Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

8 0

3 years ago

Calcular la velocidad tangencial y la velocidad angular de un movil que describe una circunferencia de 18 centimetros de radio e

Hoochie [10]

Answer:

1.41 m/s, 7.85 rad/s

Explanation:

We can start by calculating the tangential velocity, which is given by:

$v=\frac{2\pi r}{T}$

where

r = 18 cm = 0.18 m is the radius of the circumference

T = 0.8 s is the period

Substituting,

$v=\frac{2\pi(0.18)}{0.8}=1.41 m/s$

Now we can also calculate the angular velocity, which is given by:

$\omega=\frac{2\pi}{T}$

where again,

T = 0.8 s is the period

Substituting,

$\omega=\frac{2\pi}{0.8}=7.85 rad/s$

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3 years ago

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lina2011 [118]

Answer:

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3 years ago

A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact

N76 [4]

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

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3 years ago

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sertanlavr [38]

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