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ICE Princess25 [194]
3 years ago
8

Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object

's acceleration has a magnitude of 0.723 m/s2. However, when points due east and points due west, the acceleration is 0.493 m/s2, due east. Find (a) the magnitude of and (b) the magnitude of .
Physics
1 answer:
ANEK [815]3 years ago
6 0

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem.  Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them.  So let's call them F1 and F2, with F1 arger than F2.  Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.  

Can you solve this system of equations seeing them like this, or do you need more help?

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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
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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

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