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3 years ago

Elements that are gases, brittle and poor conductors are ?

Physics

1 answer:

Vinvika [58]3 years ago

6 0

Non reusable I'm not sure if this is 100% right

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The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

bonufazy [111]

Answer:

10000 V

0.00225988700565 m²

$8\times 10^{-12}\ F$

Explanation:

E = Electric field = $4\times 10^6\ V/m$

d = Gap = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Potential difference is given by

$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference between the plates is 10000 V

Area is given by

$A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2$

The area of the plate is 0.00225988700565 m²

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

The capacitance is $8\times 10^{-12}\ F$

4 0

3 years ago

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago

Please helppppppp!!!!!!!!!!!!!!!!!!!!!!!!

kykrilka [37]

Answer:

I think the Bulb No. 2 will stop emitting light if the bulb No. 1 burns out.

3 0

3 years ago

A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is

Crazy boy [7]

Answer:

$v=1.93m/s$

Explanation:

From the concept of fluids mechanics we know that if a tank has a hole at the bottom, the equation that we need to use is:

$v=\sqrt{2gh}$

Since we know gravity and its hight

$v=\sqrt{2*(9.81m/s^{2})(0.19m) }=1.93m/s$

5 0

3 years ago

How many ohms of resistance must be present in a circuit that has 240 volts and a current of 15 amps?

Mkey [24]

THE ANSWER IS 16 ohms or however its spelled

8 0

3 years ago

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