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enot [183]
3 years ago
14

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates

is not to exceed 3.00×104 V/m?
Physics
1 answer:
34kurt3 years ago
3 0

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

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A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti
LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision = 7.668\ ms^-1

Their common speed after the collision = 2.56\ ms^-1

Height achieved by the package of mass m when it rebounds = 0.33\ m

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

where K is Kinetic energy and U is Potential energy.

K= \frac{mv^2}{2} and U= mgh

Considering the fact  K_{initial} = 0\ and U_{final} =0 we will plug out he values of the given terms.

So V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1

Keypoints:

  • Sum of energies and momentum are conserved in all collisions.
  • Sum of KE and PE is also known as Mechanical energy.
  • Only KE is conserved for elastic collision.
  • for elastic collison we have e=1 that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where V_1{i} =7.668\ ms^-1.

Plugging the values we have

m\times 7.668 = (3m)\times V_{f}

Cancelling m from both sides and dividing 3 on both sides.

V_f = 2.56\ ms^-1

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of V_{f} for m mass.

As here V_{f}=-2.56\ ms^-1 we can use that if both are moving in right ward with 2.56 then there is a  -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object 1.

So

V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1

Now using law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh

\frac{v(f1)^2}{2g} = h

h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed =7.668\ ms^-1 for part b we have their common speed =2.56\ ms^-1 and for part c we have the rebound height =0.33\ m.

3 0
3 years ago
Use the drop-down menus to identify the parts of DNA. Label A: Label B:
love history [14]

Label A:

Nitrogen bases

Label B:

Sugar-phosphate backbone

7 0
3 years ago
Read 2 more answers
Do you think toxicity is a qualitative or quantitative property? explain
Tomtit [17]

Answer:

<u>Toxicity is a quantitative property</u>

Explanation:

  • Qualitative property of a object cannot be measured it can just be observed
  • Quantitative property of a substance  can be measured and be assigned a numerical value .
  • <u>The toxicity level of a substance can be measured and be assigned a numeral value </u>

<u />

.

3 0
3 years ago
The star Antares has an apparent magnitude of 1.0, whereas the star Procyon has an apparent magnitude of 0.4. Which star appears
Anika [276]

Answer:

Procyon appears brighter in the sky

Explanation:

Apparent magnitude of star Antares = 1.0

Apparent magnitude of star Procyon = 0.4

Pogson's Ratio

m₂-m₁ =  -2.50 log(B₂/B₁)

where, m is the apparent magnitude

B = Brightness of star or flux coming towards us (W/m²)

∴Larger magnitudes correspond to fainter stars so here 1>0.4 which means Antares appears dimmer and Procyon appears brighter.

6 0
3 years ago
If your car tachometer says your engine is moving at 1200 RPM, then what is it's angular velocity in Rad/sec?
Ostrovityanka [42]

Answer:

125.66 R/s

Explanation:

First    1200  r / min = 20 r/sec

20 r/s  *  2pi Radians / r = 40 pi Radians / sec = 125.66 R/s

3 0
2 years ago
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