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Lapatulllka [165]

4 years ago

11

determine the potential difference between two charged parallel plates that are .10 cm apart and have an electric field strength

of 12V/cm

1 answer:

mote1985 [20]4 years ago

6 0

The potential difference between two parallel plates is related to the electric field intensity by
$\Delta V = Ed$
where
$\Delta V$ is the potential difference
E is the electric field intensity
d is the distance between the two plates.

In our problem,
$d=0.10 cm$
and
$E=12 V/cm$
so we can calculate the potential difference as
$\Delta V= Ed = (12 V/cm)(0.10 cm)=1.2 V$

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Bas_tet [7]

Answer:

(a) Distance traveled = 75.3846 m

(b) Velocity of car at that instant will be 14 m/sec

Explanation:

We have given acceleration of the car $a=1.3m/sec^2$

Initial velocity of the cart u = 0 m/sec

(a) According to second equation of motion we know that $s=ut+\frac{1}{2}at^2$

So distance traveled by car $s_c=0\times t+\frac{1}{2}\times 1.3t^2=0.65t^2$

As the truck is moving with constant speed

So distance traveled by truck $s_t=ut=7t$

As the truck overtakes the car

So $s_c=s_t$

$0.65t^2=7t$

$t=10.769sec$

So distance traveled $s_c=s_t=7\times 10.769=75.3846m$

(b) From second equation of motion we know that v = u+at

So v = 0+1.3×10.769 = 14 m /sec

7 0

3 years ago

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitationa

sdas [7]

Answer:

The correct answer is option B)

Explanation:

Considering the given question as -

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitational pull on the shuttle? A) The moon pulls more on the shuttle. B) The earth pulls more on the shuttle. C) Both are equal due to equal distances. D) Both are equal due to the mass of the shuttle.

We know that gravitational pull (F) between any two bodies of mass $M_{1}$ and $M_{2}$ is given by -

F = $\dfrac{GM_{1}M_{2} }{r^{2} }$ where 'r' is the distance between the two bodies.

Let ,

$M_{e}$ : Mass of the earth

$M_{m}$ : Mass of the moon

m : Mass of the satellite

$r_{e}$ : Distance of satellite from earth

$r_{m}$ : Distance of satellite from moon

Given that $r_{e}$ = $r_{m}$

Let $r_{e}$ = $r_{m}$ =r

Force on satellite by the earth is -

$F_{e}$ = $\dfrac{GM_{e}m }{r^{2} }$

Force on satellite by the moon is -

$F_{m}$ = $\dfrac{GM_{m}m }{r^{2} }$

∵ Mass of earth ( $M_{e}$ ) > Mass of moon ( $M_{m}$ )

∴ $F_{e}$ > $F_{m}$

∴ The gravitational pull of earth on satellite is more than that of the moon.

4 0

4 years ago

Light of wavelength 633 nm falls on a single slit 0.5 mm wide and forms a diffraction pattern on a screen 1.0 m away from the sl

andrey2020 [161]

Answer:

The position of the first dark spot on the positive side of the central maximum is 1.26 mm.

Explanation:

Given that,

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For destructive interference :

$\dfrac{dY}{D}=n\lambda$

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Here, n = 1

$Y=\dfrac{n\lambda D}{d}\\\\Y=\dfrac{1\times 633\times 10^{-9}\times 1}{0.5\times 10^{-3}}\\\\Y=0.00126\ m\\\\Y=1.26\times 10^{-3}\ m\\\\Y=1.26\ mm$

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4 0

3 years ago

See attachment for the question

Zarrin [17]

The acceleration of the body is provided by the tension in the rope.

<h3>What is centripetal acceleration?</h3>

The centripetal acceleration is given by a = v^2/r. v = velocity of the body, r = radius

a = (8.40 m/s)^2/(8.50 m)

a = 8.3 m/s^2

The tension in the rope is the force that provides the centripetal force in the rope.

Learn more of centripetal acceleration:brainly.com/question/14465119

#SPJ1

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2 years ago

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Sophie [7]

Answer:

option D

Explanation:

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3 years ago