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2 years ago

6

16. Timothy is connecting an ungrounded

1 answer:

Mumz [18]2 years ago

5 0

We can assume that the job Timothy is working on is in the older building. Thus, the correct option is A.

<h3>What is an ungrounded receptacle?</h3>

An ungrounded receptacle may be defined as the ground wire required to deliver the same level of protection from house fires and electrical shocks suggested by grounded doorways.

The ungrounded receptacles are only furnished in the building that is already built up to a certain level with appropriate dimensions. It connects all the electrical appliances of the home through a cage of metallic or copper wire.

Therefore, the correct option is A.

To learn more about Ungrounded receptacles, refer to the link:

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A tiny, glass sphere has a charge of +10.9 nC. A metal ball bearing, which is 4.00 cm above the glass sphere, experiences an ele

andrew11 [14]

Answer:

2.12 × 10⁻⁷ C

Explanation:

Given:

Charge on the glass sphere, Q = 10.9 nC = 10.9 × 10⁻⁹ C

Distance between the glass and the metal sphere, r = 4.0 cm = 0.04 m

Force experienced by the metal sphere, F = 0.0130 N

let the charge on the metal sphere be 'q'

Now,

the force due to a charge on the other charge is given as:

$F=\frac{kQq}{r^2}$

where,

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on substituting the values in the formula, we get

$0.0130=\frac{9\times10^9\times10.9\times10^{-9}\times\q}{0.04^2}$

or

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3 years ago

A force is specified by the vector F= 160i + 80j + 120k N. Calculate the angles made by F with the positive x-, y-, and z-axis.

Rashid [163]

Answer:

1) Angle with x-axis = 42.03 degrees

2) Angle with y-axis =68.2 degrees

3) Angle with z-axis = 56.14 degrees

Explanation:

given any vector $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$

and any x axis the angle between them is given by

$\theta_x =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{i}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{x\cdot i}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_x=cos^{-1}(\frac{160}{\sqrt{160^2+80^2+120^2}} )=42.03^{o}$

Angle between the vector and y axis is given by

$\theta_y =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{j}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_y=cos^{-1}(\frac{y\cdot j}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_x=cos^{-1}(\frac{80}{\sqrt{160^2+80^2+120^2}} )=68.2^{o}$

Similarly angle between z axis and the vector is given by

$\theta_z =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{k}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{z\cdot k}{\sqrt{x^2+y^2+z^2}} )$

Applying values we get

$\theta_z=cos^{-1}(\frac{120}{\sqrt{160^2+80^2+120^2}} )=56.145^{o}$

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