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3 years ago

The total capacitance of two 15uF capacitors connected in parallel is

Physics

1 answer:

Yuri [45]3 years ago

7 0

The total capacitance of two 15uF capacitors connected in parallel is 30 μF .

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Two very quick questions!!

stepladder [879]

The weight lifted by a machine to the applied force on a machine is called mechanical advantage. This is written as Mechanical advantage, M. A, = load(weight)/effort. So for 1) M.A = 2 and load = 2, 000lb = 8896.446N. So 2 = 8896.446/ effort Effort = 8896.446/2 = 4448.48 Similarly for M.A of 2, 000, 000 we have Effort = 8896.446/ 2, 000, 000 = 0.004448

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3 years ago

Read 2 more answers

A steeper incline plane will require _____ Force

ExtremeBDS [4]

Answer:

wind

Explanation:

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3 years ago

A ball of mass 24.1 g is attached to a cord of length 0.417 m and rotates in a vertical circle. What is the minimum speed the ba

Karo-lina-s [1.5K]

Answer:

the minimum speed that the ball must have so that the cord does not become slack is 2.02 m/s.

Explanation:

In order to avoid slack, the centripetal force of the ball must equal its weight at the top of the circle. Therefore,

F_c = F_g

m v² / r = m g

v² = g r

v = √[g r]

v = √[(9.8 m/s²)(0.417 m)]

v = 2.02 m/s

Therefore, the minimum speed that the ball must have so that the cord does not become slack is 2.02 m/s.

6 0

3 years ago

Under electrostatic conditions, the electric field just outside the surface of any charged conductor:

lesantik [10]

Answer:

D. is always perpendicular to the surface of the conductor

Explanation:

1) Answer is (D) option. Electric field just outside surface of charged conductor is normal to conductor at that point.

It can be explained on the basis of the fact that, Electric field inside conductor under static condition is zero. As a result potential difference between any two points with in conductor is zero. So whole of conductor is equipotential body.

Equipotential surface and Electric field lines always cut at 90 degrees to each other. Conductor being equipotential body, Electric field lines starting or terminating at conductor must be normal to surface. Hence electric field just outside conductor is perpendicular or normal to surface.

6 0

3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago