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rewona [7]
3 years ago
9

4. What determines which goods a country should produce and export?

Physics
1 answer:
tangare [24]3 years ago
8 0

Answer:

net exporter

Explanation:

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Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charg
VashaNatasha [74]

Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charge density of 49.0 nC/m2. A proton is released from rest at the positive plate.

(a) Determine the magnitude of the electric field between the plates from the charge density.

b) Determine the potential difference between the plates.

Answer:

a

The Electric Field between the two plate is   E = 5.536*10^3 N/C

b

The potential difference between the plate is V = 609 \ Volts

Explanation:

From the we are given that

        The separation between the plate is  d =  11.0 cm

        The surface charge density is  \sigma   = 49.0 n C/m^2 = 49.0*10^{-9}

Generally Electric field between the plate is mathematically given as

                   E = \frac{\sigma }{\epsilon_0}

Note that \epsilon_0 is the permitivity of free space and its value is  8.85 *10 ^{-12} C^2 /N \cdot m^2

    Now substituting values we have

                      E = \frac{49*10^{-9}}{8.85*10^{-12}}

                        5.537*10^3 N/C

Generally Potential difference between the plate is mathematically given as

                             V =Ed      

Where E is the electric field  which is  E = 5.536*10^3 N/C

                              Substituting value we have

                           V = = (5.537*10^3 N/C) (11.0*10^{-2}) =609 V                  

 

3 0
4 years ago
after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF
Snezhnost [94]

Answer:

The time is 110.16\times10^{-3}\ sec

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

i_{0}=\dfrac{V_{0}}{R}

Put the value into the formula

i_{0}=\dfrac{150}{1.8\times10^{3}}

i_{0}=83.3\times10^{-3}\ A

We need to calculate the time

Using formula of current

i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}

Put the value into the formula

50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}

\dfrac{50}{83.3}=e^{\frac{-t}{RC}}

\dfrac{-t}{RC}=ln(0.600)

t=0.51\times1.8\times10^{3}\times120\times10^{-6}

t=110.16\times10^{-3}\ sec

Hence, The time is 110.16\times10^{-3}\ sec

4 0
3 years ago
What is the average velocity of the particle from rest to 9 seconds?
geniusboy [140]
Answer: v = 2[m/s]Explanation:This avarage velocity can be found with the ... B. 2 meters/ second. C. 3 meters/second. D. 4 meters/second. 1.
7 0
3 years ago
A very long wire carries a uniform linear charge density of 7.0 nC/m. What is the electric field strength 16.0 m from the
jok3333 [9.3K]

Electric field due to a long wire is given by

E = \frac{2 k \lambda}{r}

here

k =  9 * 10^9

\lambda = 7*10^{-9}

r = 16 m

E = \frac{2*9*10^9 *7*10^{-9}}{16}

E = 7.875 N/C

6 0
4 years ago
Help plz i have until 4.20 plz
Setler [38]

Answer:

Substance 1 because it became a liquid faster

Explanation:

8 0
3 years ago
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