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3 years ago

What is electrical ground and why is it important?

Physics

1 answer:

Likurg_2 [28]3 years ago

7 0

Answer:

One of the most important reasons for grounding electrical currents is that it protects your appliances, your home and everyone in it from surges in electricity. If lightning was to strike or the power was to surge at your place for whatever reason, this produces dangerously high voltages of electricity in your system.

Explanation:

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A 3.50 kg box that is several hundred meters above the surface of the earth is suspended from the end of a short vertical rope o

Anni [7]

Answer:

a. i. -4.65 m/s ii. -13.95 m/s b. 5.89 m c. 2.85 s

Explanation:

a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?

We write the equation of the forces acting on the mass.

So, T - mg = ma where T = tension in vertical rope = (36.0 N/s)t, m = mass of box = 3.50 kg, g = acceleration due to gravity = 9.8 m/s² and a = acceleration of box = dv/dt where v = velocity of box and t = time.

So, T - mg = ma

T/m - g = a

dv/dt = T/m - g

dv/dt = (36.0 N/s)t/3.50 kg - 9.8 m/s²

dv/dt = (10.3 m/s²)t - 9.8 m/s²

dv = [(10.3 m/s²)t - 9.8 m/s²]dt

Integrating, we have

∫dv = ∫[(10.3 m/s³)t - 9.8 m/s²]dt

∫dv = ∫(10.3 m/s³)tdt - ∫(9.8 m/s²)dt

v = (10.3 m/s³)t²/2 - (9.8 m/s²)t + C

v = (5.15 m/s³)t² - (9.8 m/s²)t + C

when t = 0, v = 0 (since at t = 0, box is at rest)

So,

0 = (5.15 m/s³)(0)² - (9.8 m/s²)(0) + C

0 = 0 + 0 + C

C = 0

So, v = (5.15 m/s³)t² - (9.8 m/s²)t

i. What is the velocity of the box at t = 1.00 s,

v = (5.15 m/s³)(1.00 s)² - (9.8 m/s²)(1.00 s)

v = 5.15 m/s - 9.8 m/s

v = -4.65 m/s

ii. What is the velocity of the box at t = 3.00 s,

v = (5.15 m/s³)(3.00 s)² - (9.8 m/s²)(3.00 s)

v = 15.45 m/s - 29.4 m/s

v = -13.95 m/s

b. What is the maximum distance that the box descends below its initial position?

Since v = (5.15 m/s³)t² - (9.8 m/s²)t and dy/dt = v where y = vertical distance moved by mass and t = time, we need to find y.

dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t

dy = [(5.15 m/s³)t² - (9.8 m/s²)t]dt

Integrating, we have

∫dy = ∫[(5.15 m/s³)t² - (9.8 m/s²)t]dt

∫dy = ∫(5.15 m/s³)t²dt - ∫(9.8 m/s²)tdt

∫dy = ∫(5.15 m/s³)t³/3dt - ∫(9.8 m/s²)t²/2dt

y = (1.72 m/s³)t³ - (4.9 m/s²)t² + C'

when t = 0, y = 0.

So,

0 = (1.72 m/s³)(0)³ - (4.9 m/s²)(0)² + C'

0 = 0 + 0 + C'

C' = 0

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

The maximum distance is obtained at the time when v = dy/dt = 0.

So,

dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t = 0

(5.15 m/s³)t² - (9.8 m/s²)t = 0

t[(5.15 m/s³)t - (9.8 m/s²)] = 0

t = 0 or [(5.15 m/s³)t - (9.8 m/s²)] = 0

t = 0 or (5.15 m/s³)t = (9.8 m/s²)

t = 0 or t = (9.8 m/s²)/(5.15 m/s³)

t = 0 or t = 1.9 s

Substituting t = 1.9 s into y, we have

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

y = (1.72 m/s³)(1.9 s)³ - (4.9 m/s²)(1.9 s)²

y = (1.72 m/s³)(6.859 s³) - (4.9 m/s²)(3.61 s²)

y = 11.798 m - 17.689 m

y = -5.891 m

y ≅ - 5.89 m

So, the maximum distance that the box descends below its initial position is 5.89 m

c. At what value of t does the box return to its initial position?

The box returns to its original position when y = 0. So

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

0 = (1.72 m/s³)t³ - (4.9 m/s²)t²

(1.72 m/s³)t³ - (4.9 m/s²)t² = 0

t²[(1.72 m/s³)t - (4.9 m/s²)] = 0

t² = 0 or (1.72 m/s³)t - (4.9 m/s²) = 0

t = √0 or (1.72 m/s³)t = (4.9 m/s²)

t = 0 or t = (4.9 m/s²)/(1.72 m/s³)

t = 0 or t = 2.85 s

So, the box returns to its original position when t = 2.85 s

6 0

3 years ago

Describe what introspection is.

krek1111 [17]

Answer:

Introspection is a process that involves looking inward to examine one's own thoughts and emotions. ... The experimental use of introspection is similar to what you might do when you analyze your own thoughts and feelings but in a much more structured and rigorous way.

Explanation:

Example of it: The definition of introspection is self-examination, analyzing yourself, looking at your own personality and actions, and considering your own motivations. An example of introspection is when you meditate to try to understand your feelings. noun.

6 0

3 years ago

Read 2 more answers

Which of the following best describes chemical weathering?

dexar [7]

Answer:

<h3>ir id n becuswe ig gyst makres sensre</h3>

Explanation:

5 0

3 years ago

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Which are advantages of reflecting telescopes? Check all that apply.

Mashcka [7]

On e2020 the answers are:

1. There is no rainbow-like halo around the image.

2. Reflecting telescopes can be made very large.

4. Only the reflecting side of the primary mirror needs to be perfectly shaped and smooth.

I hope that helps!!

3 0

3 years ago

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An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

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4 years ago