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3 years ago

Please help on this one?

Physics

2 answers:

jekas [21]3 years ago

8 0

It should be A) Force increases

stira [4]3 years ago

7 0

Density decreases, because what was once big, turned small, thus, there is a decrease in density.

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago

Which of the following is incorrect for nuclear forces

densk [106]

Answer:

c they obey inverse square law

4 0

3 years ago

Read 2 more answers

A model rocket is launched with an initial upward velocity of.

REY [17]

Answer:
A model rocket is launched with an initial upward velocity of 215 ft/s.

Explanation:

3 0

2 years ago

How does the current in a circuit change if the resistance is doubled? The current is halved. The current is doubled. The curren

Inessa05 [86]

Answer:

The current is halved

Explanation:

The relationship between the current and the resistance is given by Ohm's Law, as follows:

$V = IR\\I = \frac{V}{R}$

where,

V = Voltage

I = Current

R = Resistance

Therefore, if we double the resistance:

$I' = \frac{V}{2R}\\\\I' = \frac{1}{2}I$

Hence the correct option is:

<u>The current is halved</u>

7 0

3 years ago

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A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th

vladimir1956 [14]

Answer:

Force, $F=2.27\times 10^4\ N$

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

$F.d=\dfrac{1}{2}m(v^2-u^2)$

Finally, it stops, v = 0

$F.d=-\dfrac{1}{2}m(u^2)$

$F=\dfrac{-mu^2}{2d}$

$F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}$

F = -22713.92 N

$F=2.27\times 10^4\ N$

So, the magnitude of the force that stops the bullet is $2.27\times 10^4\ N$

7 0

3 years ago