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baherus [9]
3 years ago
15

A 16.9 kg monkey is swinging on a 5.32 m long vine. It starts at rest, with the vine at a 43.0° angle. How fast is the monkey mo

ving at the bottom of its swing? (Unit = m/s) ​
Physics
1 answer:
NemiM [27]3 years ago
3 0

Answer:

v = 5.7554 m/s

Explanation:

First of all we need to know if the angle of the vine is measured in the horizontal or vertical.

To do this easier, let's assume the angle is measured with the horizontal. In this case, the innitial height of the monkey will be:

h₀ = h sinα

h₀ = 5.32 sin43° = 3.6282 m

As the monkey is dropping from the innitial point which is the suspension point, is also dropping from 5.32. Then the actual height of the monkey will be:

Δh = 5.32 - 3.63 = 1.69 m

In order to calculate the speed of the monkey we need to understand that the monkey has a potential energy. This energy, because of the gravity, is converted in kinetic energy, and the value will be the same. Therefore we can say that:

Ep = Ek

From here, we can calculate the speed of the monkey.

Ep = mgΔH

Ek = 1/2 mv²

The potential energy is:

Ep = 16.9 * 9.8 * 1.69 = 279.9

Now with the kinetic energy:

1/2 * (16.9) * v² = 279.9

v² = (279.9) * 2 / 16.9

v² = 33.12

v = √33.12

<h2>v = 5.7554 m/s</h2>

Hope this helps

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Answer:

a. 6.032\times10^{-6}C/m^2

b.6.816\times10^5N/C

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

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Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2  #surface charge outside sphere.

\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2

Hence, the new charge density on the outside of the sphere is 6.032\times10^{-6}C/m^2

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be 6.032\times10^{-6}C/m^2,

E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C

Hence, the strength of the electric field just outside the sphere is 6.816\times10^5N/C

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