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juin [17]
3 years ago
7

Problem 21-40a:

Physics
1 answer:
Greeley [361]3 years ago
8 0

Answer:

Part a)

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

q = 3.37 \times 10^{-4} C

Explanation:

As we know that electric force on electric charge is given as

F = qE

here we have

q = 1.6 \times 10^{-19}C

E = 153 N/C

now force is given as

F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N

Gravitational force on electric charge near surface of earth is given as

F_g = mg

F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N

now the ratio of two forces is given as

\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

F_g = qE

mg = qE

(5.25 \times 10^{-3})(9.81) = q(153)

here we have

q = 3.37 \times 10^{-4} C

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Answer:

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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}

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r=\frac{4}{5}h

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

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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:

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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}}

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

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fredd [130]

Answer:

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Explanation:

<u>Uniform Acceleration </u>

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The relation between the initial and final speeds is:

v_f=v_o+a.t

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It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.

Solving for a:

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Substituting:

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