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malfutka [58]
4 years ago
9

What is the total energy equation?

Physics
1 answer:
netineya [11]4 years ago
6 0
The total energy equation would be Kinetic energy+Potential energy
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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
If 710- nm and 660- nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wa
alexira [117]

0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.

<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.5m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3276.92 x 10⁻⁶ m

= 3.276x 10⁻³ m

= 3.276mm .

For λ = 660 nm

position = 2 λ D / d

λ = 660 nm , D = 1.5 m

d = .65 x 10⁻³

position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3046.15 x 10⁻⁶ m

= 3.046 x 10⁻³ m

= 3.046 mm .

Difference between their position

= 3.276mm ₋ 3.046 mm

= 0.23 mm .

To know more about Fringes refer to:  brainly.com/question/15649748

#SPJ4

7 0
2 years ago
Please Help with this
Tresset [83]

Answer: c is correct

Explanation: i did this

3 0
3 years ago
In calisthenics, the resistance is ultimately provided by __________. A. free weights B. springs and elastic bands C. gravity D.
tangare [24]

Answer:

I think the answer is b am sorry if it is wrong

Explanation:

5 0
3 years ago
Read 2 more answers
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon
LiRa [457]

Answer:

241.8 N.

Explanation:

The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.

Centripetal force = mv^2/r

F = mg + mv²/r

F = m(g + v²/r)

where,

m = mass

= 9 kg

g = acceleration due to gravity

= 9.8 m/s²

v = 3.2 m/s

r = 0.60 m

F = 9 * (9.8 + 3.2²/0.60)

= 241.8 N.

3 0
3 years ago
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