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2 years ago

To prevent injury while maintaining equipment make sure the equipment is

1 answer:

3 0

To prevent injury while maintaining equipment make sure the equipment is locked out and/or tagged out (i.e., de-energized).

<h3>What does locked-out equipment mean?</h3>

Equipment is locked out when it does not contain any type of change that may affect the security of a device.

This task (locked out) is fundamental to avoid any safety problems associated with a device.

In conclusion, to prevent injury while maintaining equipment make sure the equipment is locked out and/or tagged out (i.e., de-energized).

Learn more about locked-out equipment here:

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What two items were combined to launch the film canister rocket?

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A point charge q1 = -4. 00 nc is at the point x = 0. 60 m, y = 0. 80 m , and a second point charge q2 = +6. 00 nc is at the poin

Alekssandra [29.7K]

The net electric field is the vector sum of the components of the electric

field produced by the two charges.

The values of the magnitude and direction of the net electric field at the origin (approximate values) are;

131.6 N/C

12.6 ° above the negative x–axis

<h3>How are the net electric field magnitude and direction calculated?</h3>

The possible questions based on a similar question posted online are;

(a) The net electric field at the origin.

The electric field due to charge q₁ is given as follows;

$\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}$

Which gives;

$\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C$

$\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C$

Which gives;

$\vec{E}_1 = \mathbf{21.6 \, N/C \cdot \hat x + 28.8 \, N/C \hat y}$

$\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C$

Therefore;

$\vec {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)$

$\vec {E} = \mathbf{\left( -128.4 \, N/C \right) \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)}$

The magnitude of the net electric field is therefore;

E = $\sqrt{(-128.4^2 + 28.8^2)}$ ≈ 131.6

The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>

(b) The direction of the net electric field at the origin.

$The \ direction \ is \ arctan \left(\dfrac{28.8}{-128.4} \right) \approx \underline{ 12.6^{\circ}} \ above \ the \ negative \ x-axis$

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2 years ago

A humpback whale is classified as a mammal while a basking shark is a fish.

kvasek [131]

The taxonomic classification that is the same for a humpback whale and a basking shark is Phylum.

<h3>What is Taxonomic Classification?</h3>

We want to find Which taxonomy classification are the same for a humpback whale and a basking shark.

A chordate in the animal kingdom is referred to as Kingdom Animalia of the phylum Chordata. This phylum is one that includes all the vertebrates and subphyla tunicates and cephalochordates.

Chordates have peculiar characteristics such as notochord, pharyngeal slits, nerve cord, and a tail.

Therefore, we can conclude that the taxonomic classification that is the same for a humpback whale and a basking shark is Phylum.

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This app is life saving!! <br> I love this app <br> Btw can someone give me extra point

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