Ask question

Ask question

All categories

3 years ago

14

Suppose you had a point charge of 3.6 mC. show answer Incorrect Answer Calculate the magnitude of the electric field, in newtons

per coulomb, at a point 1.4 m away from this charge.

1 answer:

Ksivusya [100]3 years ago

5 0

Answer: 1.653 × 10⁴ N/C

Explanation: Given Q= 3.6× 10^-6

we assume there is a charge q placed at a distance r = 1.4m exerting an electric field strength towards Q.

Hence

F = kQq/r²

K= 9× 10^9

Now recall that E = F/q (electric field due to that charge)

E =( kQq/r²)/q

E = kQ/r² substituting the values gave the answer above.

You might be interested in

An electron and a proton are each placed at rest in a uniform electric field of magnitude 560 N/C. Calculate the speed of each p

Butoxors [25]

Answer:

The speed of electron is $v=4.52\times 10^6\ m/s$ and the speed of proton is 2468.02 m/s.

Explanation:

Given that,

Electric field, E = 560 N/C

To find,

The speed of each particle (electrons and proton) 46.0 ns after being released.

Solution,

For electron,

The electric force is given by :

$F=qE$

$F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N$

Let v is the speed of electron. It can be calculated using first equation of motion as :

$v=u+at$

u = 0 (at rest)

$v=\dfrac{F}{m}t$

$v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}$

$v=4.52\times 10^6\ m/s$

For proton,

The electric force is given by :

$F=qE$

$F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N$

Let v is the speed of electron. It can be calculated using first equation of motion as :

$v=u+at$

u = 0 (at rest)

$v=\dfrac{F}{m}t$

$v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}$

$v=2468.02\ m/s$

So, the speed of electron is $v=4.52\times 10^6\ m/s$ and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.

6 0

4 years ago

How many significant figures?<br> 5.0001<br> O None of these are correct<br> O 5<br> 02<br> 0 1

mezya [45]

5

if zero falls between two significant numbers it becomes significant.

6 0

4 years ago

A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 15

Sveta_85 [38]

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

$V_{A}$ = 1 m³

$T_{A}$ = 10°C = 283 K

$P_{A}$ = 350 kPa

$m_{B}$ = 3 kg

$T_{B}$ = 35°C = 308 K

$P_{B}$ = 150 kPa

Now, lets apply the ideal gas equation;

$P_{B}$ $V_{B}$ = $m_{B}$ R $T_{B}$

$V_{B}$ = $m_{B}$ R $T_{B}$ / $P_{B}$

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

$V_{B}$ = ( 3 × 0.287 × 308) / 150

$V_{B}$ = 265.188 / 150

$V_{B}$ = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, $m_{A}$ = $P_{A}$ $V_{A}$ / R $T_{A}$ = (350 × 1)/(0.287 × 283) = 350 / 81.221

$m_{A}$ = 4.309 kg

Total mass, $m_{f}$ = $m_{A}$ + $m_{B}$ = 4.309 + 3 = 7.309 kg

Total volume $V_{f}$ = $V_{A}$ + $V_{B}$ = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

$P_{f}$ = $m_{f}$ R $T_{f}$ / $V_{f}$

given that; final temperature $T_{f}$ = 20°C = 293 K

we substitute

$P_{f}$ = ( 7.309 × 0.287 × 293) / 2.77

$P_{f}$ = 614.6211119 / 2.77

$P_{f}$ = 221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

6 0

3 years ago

Which best describes the way a sound wave is sent through the radio?

Anna35 [415]

Answer:

sound wave-electrical wave-radio wave

4 0

3 years ago

Read 2 more answers

How is electromagnetic energy from the microwave transformed into heat energy

8_murik_8 [283]

The energy comes in waves and I forget how, but the waves make the water molecules in teh food spin and that causes friction which warms the food up

7 0

3 years ago

Read 2 more answers