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enot [183]
4 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:
34kurt4 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 motor spins up the flywheel with a constant torque of 50 n⋅m. how long does it take the flywheel to reach top speed?
dimulka [17.4K]

A motor spins upward the flywheel with a persistent torque of 50N⋅m.

What time does it take the flywheel to get to the top speed?
From the equation: 
Tj = J*dω/dt 


you can get the two equations: 
Δt1= J1*Δω/Tj = 240*125.66/50 = 603.17 sec 
Δt2= J2*Δω/Tj = 120*125.66/50 = 301.58 sec 

6 0
4 years ago
10 The magnitude J of the current density in a certain lab wire with a circular cross section of radius R = 2.00 mm is given by
rewona [7]

Answer:

I = 0.002593 A = 2.593 mA

Explanation:

Current density = J = (3.00 × 10⁸)r² = Br²

B = (3.00 × 10⁸) (for ease of calculations)

The current through outer section is given by

I = ∫ J dA

The elemental Area for the wire,

dA = 2πr dr

I = ∫ Br² (2πr dr)

I = ∫ 2Bπ r³ dr

I = 2Bπ ∫ r³ dr

I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]

I = (Bπ/2) [R⁴ - (0.9R)⁴]

I = (Bπ/2) [R⁴ - 0.6561R⁴]

I = (Bπ/2) (0.3439R⁴)

I = (Bπ) (0.17195R⁴)

Recall B = (3.00 × 10⁸)

R = 2.00 mm = 0.002 m

I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]

I = 0.0025929449 A = 0.002593 A = 2.593 mA

Hope this Helps!!!

4 0
3 years ago
A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,
Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a)  Kinetic energy of block = potential energy in spring  

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg.  Spring constant k is unknown, but you can find it from given data:  

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.  

From the energy equation above, solve for v,

v = v √(k/m)  

= 0.15 √(300/1)

= 2.598 m/s.

b)  Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.  

This is the same momentum carried by bullet as it strikes the block.  Therefore, if u is bullet speed,  

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0
3 years ago
What is one benefit of lifelong physical activity?
kvasek [131]
The answer would be C
5 0
3 years ago
If bert the bat traveles eastward at 40 mph with a tail wind of 6 mph,what is his actual speed?
Liula [17]
Bert would be going 34mph due to the taill winds being 6mph
5 0
3 years ago
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