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liraira [26]
2 years ago
12

A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent

ial difference would appear as:_________
Physics
1 answer:
vfiekz [6]2 years ago
5 0

The potential difference comes out to be

10 \times 10 {}^{ - 3} m

Given:

σ = 8. 85 × 10-9 c/m2

we know,

E = \frac{σ}{2ε0}

E =  \frac{8.85 \times 10 {}^{ - 9} }{2ε0}

E =  \frac{v}{d}

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

=  \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }

Δ = 10 \times 10 {}^{ - 3} m

Thus the potential difference is

10 \times 10 {}^{ - 3} m

Learn more about potential difference from here: brainly.com/question/28165869

#SPJ4

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A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 ×
otez555 [7]

Answer:

t = 300.3 seconds

Explanation:

Given that,

The mass of a freight train, m=1.01\times 10^7\ kg

Force applied on the tracks, F=7.5\times 10^5\ N

Initial speed, u = 0

Final speed, v = 80 km/h = 22.3 m/s

We need to find the time taken by it to increase the speed of the train from rest.

The force acting on it is given by :

F = ma

or

F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s

So, the required time is 300.3 seconds.

4 0
3 years ago
A man pushes a 60.8 kg crate across a rough surface with an applied force of 125 N and at a CONSTANT SPEED.
Sedbober [7]

Answer: Colby we both dumb if we need brainly lol

Explanation:

8 0
4 years ago
HELP the decay curve of twizzlers sheet!!!!!
Vinvika [58]
Ok so the first thing you need to do is im not gonna help you
7 0
3 years ago
With what speed must you approach a source of sound to observe a 25% change in frequency?
insens350 [35]
Sound source is at rest, you are moving with velocity v, f = frequency, c = speed of sound:

f = f0(1 + v/c)

115 = 100(1 + v/343)
115 = 100 + 100v/343
15 = 100v/343
v = 15*343/100
<span> v = 51,45 m/s </span>
5 0
3 years ago
Stars of spectral type A and F are considered ________.
LekaFEV [45]

Answer:

<u>B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.</u>

Explanation:

The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". <em>This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K</em> (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:

• They live at least a few billion years, allowing life a chance to evolve. <em>More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.</em>

• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.

• They emit sufficient radiation at wavelengths conducive to photosynthesis.

• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.

<u><em>Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.</em></u>

4 0
3 years ago
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