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Alex787 [66]
3 years ago
9

How does a parallel circuit differ from a series circuit?

Physics
1 answer:
8_murik_8 [283]3 years ago
5 0
I think the correct answer from the choices listed above is option B. A parallel circuit differ from a series circuit in a sense that a <span>series circuit has one path for electrons, but a parallel circuit has more than one path. In a parallel circuit there two or more paths for current to flow while a series circuit only has one.</span>
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The established value for the speed of light in a vacuum is 299,792,458 m/s. What is the order-of-magnitude of this number?
iogann1982 [59]
The order of magnitude is 10⁸ .
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An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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3 years ago
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Answer:

◆ See the attachment photo.

◆ Don't forget to thanks

◆ Mark as brainlist.

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3 years ago
A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower
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Answer:

  She will make the jump.

Explanation:

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

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Substituting

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So she will cover 77 m in 2.85 seconds

 Now considering vertical motion, up direction as positive

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  So she will make the jump.

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3 years ago
You find yourself in a place that is unimaginably hot and dense. A rapidly changing gravitational field randomly warps space and
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You find yourself in a place that is unimaginably <u>hot and dense</u>. A r<u>apidly changing</u><u> gravitational field</u><u> </u>randomly warps space and time. Gripped by these huge fluctuations, you notice that there is but a single, unified force governing the universe, you are in the early universe before the Planck time.

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The Planck time is approximately<u> 10^-44 seconds</u>. The smallest time interval, or "zeptosecond," that has so far been measured is <u>10^-21 seconds</u>. A photon traveling at the speed of light would need one Planck time <u>to traverse a distance of one </u><u>Planck length</u>.

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To learn more about Plank time:

brainly.com/question/23791066

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8 0
2 years ago
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