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2 years ago

Please help on this one somebody? :)

Physics

1 answer:

ahrayia [7]2 years ago

8 0

ans D. Substance B has a greater latent heat of vaporization than substance A.

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____________ showed that matter and energy are two forms of the same thing.

Darya [45]

Einstein. That genius

7 0

2 years ago

Read 2 more answers

Correctly round the following number to the whole number<br> 2.35784

Elina [12.6K]

Answer:2

Explanation:

8 0

2 years ago

The dome of a Van de Graaff generator receives a charge of 0.00011 C. The radius of the dome is 5.2 m. Find the strength of the

wel

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

F=kq1/r²

F=kQ/r²

F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

Then Electric field is given as,

E=kQ/r²

E=8.98755E9 ×0.00011/12.6²

E=6227.34N/C

5 0

3 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

The acceleration due to gravity on the moon is about 5.4ft/s2. if your weight is 150lbf on earth

Ivenika [448]

... then your weight is <em>25.2 lbf</em> on the moon.

6 0

3 years ago