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malfutka [58]
3 years ago
12

Two point charges lie on the x axis. A charge of + 2.20 pC is at the origin, and a charge of − 4.80 pC is at x=−11.0cm. What thi

rd charge should be placed at x=+20cm so that the total electric field at x=+10.0cm is zero?
Physics
2 answers:
k0ka [10]3 years ago
4 0
As we know that electric field at any point caused due to a point charge given by, kQ/r^2

Where, k is dielectric constant of medium, Q is the point charge and r is the distance between the point charge and point where electric field is to be measured.

Let us assume a charge of q is placed at x=+20cm to make electric field at
x=+10cm zero.

Total electric field at x=+10cm can be shown by equation,

(k X 2.20 X 10^-12)/(10-0)^2  +  (k X -4.80 X 10^-12)/(10-(-11))^2  + (k X
q)/(10-20)^2  = 0

=> 2.20 X 10^-14 - 1.09 X 10^-14 + q X 10^-2 = 0

=> q = 1.09 X 10^-12 - 2.20 X 10^-12

=> q = -1.1 X 10^-12 Coulombs or -1.1 pC
My name is Ann [436]3 years ago
4 0

Answer: The charge must be -1.11pC

Explanation: The field done by a charge is equal to:

E(x) = k*q/(x - x')^2

Where k is the constant, q is the charge and x' is the position of the charge, then, the total field at the point x = 10cm must be also zero, we have:

E(10cm) = k*(2.20pC/(10cm - 0cm)^2 - 4.80pC/(10cm + 11cm)^2 + Q/(10cm - 20cm)^2) = 0

And we must find the value of Q.

(2.20pC/(10cm)^2 - 4.80pC/(21cm)^2 + Q/(-10cm)^2) = 0

Q = (-2.20pC/(10cm)^2 + 4.80pC/(21cm)^2)*(10cm)^2 = -1.11pC

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A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
3 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

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now in part a)

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so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

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Part c)

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