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Nutka1998 [239]

3 years ago

11

A projectile is fired horizontally off the top of a cliff with an initial velocity of 30 m/s. It hits the ground 2.0 seconds lat

er. How far from the base of the cliff does the projectile land?

1 answer:

Mandarinka [93]3 years ago

6 0

Answer:

The horizontal distance traveled by the projectile is 60 m

Explanation:

Given;

initial horizontal velocity of the projectile, Vₓ = 30 m/s

time of the motion of the projectile, t = 2 s

The horizontal distance traveled by the projectile is given by the range of the projection;

X = Vₓt

X = 30 x 2

X = 60 m

Therefore, the horizontal distance traveled by the projectile is 60 m

Therefoe

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A Gaussian surface in the form of a hemisphere of radius R = 3.04 cm lies in a uniform electric field of magnitude E = 1.64 N/C.

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Answer:

Part a)

$\phi = -4.76 \times 10^{-3} Nm^2/C$

Part b)

$\phi_{curved} = 4.76 \times 10^{-3} Nm^2/C$

Explanation:

Part a)

Electric flux entering into the base

so it is given as

$\phi = E.A$

$\phi = - EA$

$\phi = -(1.64)(\pi r^2)$

$\phi = -(1.64)(\pi\times 0.0304^2)$

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Part b)

Now since we know that there is no enclosed charge in the hemisphere

so net flux must be zero

$\phi_{curved} + \phi_{flat} = 0$

$\phi_{curved} - 4.76 \times 10^{-3} = 0$

$\phi_{curved} = 4.76 \times 10^{-3} Nm^2/C$

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Which substance is used in fertilizers?

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3 years ago

Read 2 more answers

lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of

bonufazy [111]

Answer:

The intensity of laser 2 is 4 times of the intensity of laser 1.

Explanation:

The intensity in terms of electric field is given by :

$U=\dfrac{1}{2}\epsilon_o E^2$

E is electric field

It means, $U\propto E^2$

In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.

Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,

$\dfrac{U}{U'}=(\dfrac{E}{E'})^2$

We have,

E'=2E

So,

$\dfrac{U}{U'}=\dfrac{E^2}{(2E)^2}\\\\\dfrac{U}{U'}=\dfrac{E^2}{4E^2}\\\\\dfrac{U}{U'}=\dfrac{1}{4}\\\\U'=4\times U$

So, the intensity of laser 2 is 4 times of the intensity of laser 1.

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3 years ago

What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?

IRISSAK [1]

Answer:

The magnitude of electric field is 22.58 N/C

Solution:

Given:

Force exerted in upward direction, $\vec{F_{up}} = 3.50\times 10^{- 5} N$

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Now, we know by Coulomb's law,

$F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}$

Also,

Electric field, $E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}$

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

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$\vec{E} = - 22.58 N/C$

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

$|\vec{E}| = 22.58\ N/C$

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3 years ago