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Igoryamba
4 years ago
9

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

se charges at a point at on the x-axis?
Physics
1 answer:
Airida [17]4 years ago
6 0

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

V =\frac{Kq}{r}

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V

Total potential due to this charges  = 4500 V + 6750 V = 11250 V

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Answer:

"13.48 Kwhr" is the right solution.

Explanation:

The given values are:

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= 0.063 W/m²

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Efficiency of conversion,

= 50%

Now,

The area of hotspot will be:

⇒  A=\frac{\pi}{4} d^2

On substituting the values, we get

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⇒      =0.785\times 64

⇒      =50.24 \ m^2

Total heat generation rate will be:

⇒  Q=q\times A

        =0.063\times 50.24

        =3.16 \ W

hence,

The electricity generation capacity will be:

⇒  P=\eta Q t

On substituting the values, we get

⇒      =0.5\times 3.16\times 8760

⇒      =13840.8 \ Whr

On converting into Kwhr, we get

⇒      =13.84 \ Kwhr

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3 years ago
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At which angle must a laser beam enter the water for no refraction to occur?
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5 0
3 years ago
2. A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.5 . At 30.0 s after blast
harina [27]

Answer:

a)The highest point reached by the rocket is 1412 m

b)The rocket crashes after 54.7 s

Explanation:

Hi there!

The equations of height and velocity of the rocket are the following:

h = h0 + v0 · t + 1/2 · a · t² (while the engines work).

h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).

v = v0 + a · t (while the engines work).

v = v0 + g · t (when the rocket is in free fall).

Where:

h = height of the rocket at a time t.

h0 = initial height of the rocket.

v0 = initial velocity.

t = time.

a = acceleration due to the engines.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the rocket at a time t.

First, let's find the velocity and height reached by the rocket until the engines fail:

h = h0 + v0 · t + 1/2 · a · t²

Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:

h = 1/2 · a · t²

h = 1/2 · 2.5 m/s² · (30.0 s)²

h = 1125 m

Now let's find the velocity of the rocket at t = 30.0 s:

v = v0 + a · t (v0 = 0)

v = 2.5 m/s² · 30.0 s

v = 75 m/s

After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:

v = v0 + g · t

0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).

-75 m/s / -9.8 m/s² = t

t = 7.7 s

Now, let's find the height of the rocket 7.7 s after the engines fail:

h = h0 + v0 · t + 1/2 · g · t²

The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:

h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²

h = 1412 m

The highest point reached by the rocket is 1412 m

b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.

h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)

0 = 1412 m - 1/2 · 9.8 m/s² · t²

-1412 m / -4.9 m/s² = t²

t = 17 s

The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².

Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².

Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²

The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s

6 0
3 years ago
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