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kvv77 [185]
4 years ago
13

The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time

Physics
1 answer:
Makovka662 [10]4 years ago
7 0

Explanation:

We have,

Surface area, A=2\ cm^2=0.0002\ m^2

The current varies wrt time t as :

q(t) = 4t^3 + 5t + 6

(a) At t = 2 seconds, electrical charge is given by :

q(t) = 4t^3 + 5t + 6\\\\q(2) = 4(2)^3 + 5(2) + 6\\\\q=48\ C

(b) Current is given by :

I=\dfrac{dq}{dt}\\\\I=\dfrac{d(4t^3 + 5t + 6)}{dt}\\\\I=12t^2+5

Instantaneous current at t = 1 s is,

I=12(1)^2+5=17\ A

(c) Current is, I=12t^2+5

Current density is given by electric current per unit area.

J=\dfrac{I}{A}\\\\J=\dfrac{(12t^2+5)}{0.0002}\\\\J=5000(12t^2+5)\ A/m^2

Therefore, it is the required explanation.

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A 150 N force acts on an object for five seconds in changes the objects velocity by 30 m/s. What is the mass of the object?
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F=ma

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3 years ago
A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su
Damm [24]

Answer:

The orbital period of the  planet would be half that of the earth

Explanation:

From the question we are told that

         The mass of star is four times the mass of sun which can be mathematically represented as

                        M_{star } = 4M_{sun}

Mathematically gravitational potential is given as

                          F = \frac{K M m}{r^2}

  Where M is mass one

               m is mass two

  From this equation we see that the attraction force is directly proportional to the mass of the star

    Thus we can say that

              F_{planet } = 4 F_{earth }

The centrifugal force that balances this attraction Force  is  

                F = a_c* m

Where  a_c is the centrifugal acceleration which can be mathematically represented as a_c = \frac{r}{T^2}

          and  m is the mass

    Substituting this into the equation for centrifugal force

                         F = \frac{r}{T^2}*m

       substituting  into the equation above

                    \frac{r}{T_2^2} *m = 4(\frac{r}{T_1^2} *m)

Given that the diameter is the same and assuming that the mass is constant

         Then

                    \frac{1}{T_2^2}  = 4(\frac{1}{T_1^2} )

                   T_2 ^2 = \frac{T_1^2}{4}

Take square root of both sides

                  T_2 = \frac{T_1}{2}

                 

7 0
4 years ago
A 500 kg object is hanging from a spring attached to the ceiling. If the spring constant in the spring is 900 N/kg, how far does
My name is Ann [436]

We are given that a 500 kg object is hanging from a spring. To determine the amount the spring is stretched we will use Hook's law, which states the following:

F=kx

Where:

\begin{gathered} F=\text{ force} \\ k=\text{ spring constant} \\ x=\text{ distance stretched} \end{gathered}

Since the object is hanging the only force acting on the spring is the weight of the object. The weight of the object is:

F_g=mg

Where:

\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}

Plugging in the values we get:

F_g=(500\operatorname{kg})(9.8\frac{m}{s^2})

Solving the operations:

F_g=4900N

Now we solve for "x" from Hook's law by dividing both sides by "k":

\frac{F}{k}=x

Now we plug in the known values:

\frac{4900N}{900\frac{N}{m}}=x

Solving the operations:

5.4m=x

Therefore, the spring is stretched by 5.4 meters.

7 0
1 year ago
explain why water has a different boiling point at an elevation of 3000 meters that it does at sea level
Scorpion4ik [409]
This is because the pressure is so much larger at 3000 meters than it is at sea level. Hope this helps.
3 0
3 years ago
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