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4 years ago

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A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density

of +6.37×10−6C/m2. A charge of −0.500μC is now introduced at the center of the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

1 answer:

frez [133]4 years ago

5 0

Answer: a)5.73 *10^-6 C/m^2; b)648*10· N/C c)56.5 *10^3 Nm^2/C

Explanation: Considering the definition of surface charge density as:

σ = Q/ area of sphere

so Q = σ * 4*π* R^2outer

The new Q at the outer radius is: Q initial -0.5 microC =5 micro C

The Electric with the new outer charge is equal to:

E= k*Q / R^2outer= 9 10^9* 5 microC/(0.25m)^2= 648*10^3 N/C

Finally the flux corresponding to the internal charge is equal to

By using Gauss law, the total flux is equal to net change divided ε0

so this is 0.5 microC/ ε0= 56.5 *10^3 Nm^2/C

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A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocit

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Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)= $1.31 m/s^2$

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

$5.47=0+1.31\times t$

$t=\frac{5.47}{1.31}=4.17 s$

Distance traveled during this time is

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$s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m$

So remaining distance left to travel with constant velocity=153.57 m

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3 years ago

A filing cabinet weighing 556 N rests on the floor. The coefficient of static friction between it and the floor is 0.68, and the

zvonat [6]

Explanation:

"Static friction is a force that keeps an object at rest. It must be overcome to start moving the object."

(556 x 0.68) = static friction of 378.08N. before movement occurs.

The forces (a) and (b) will not move it.

Each will incur a frictional force preventing movement equal to itself, = 222N. and 334N. respectively.

Forces (c) and (d) will move it, and accelerate it.

Forces (c) and (d) will both encounter friction of (556 x 0.56) = 311.36N. when the cabinet is moving.

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3 years ago

Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren

nikklg [1K]

Answer:

$4.77\ \text{A}$

Explanation:

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$I_n$ = Net current

$I_2$ = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

$\theta$ = Angle between current and magnetic field = $65^{\circ}$

$l$ = Length of wires = 2.64 m

$I$ = Current in the other wire

Magnetic force is given by

$F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}$

Net current is given by

$I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}$

The current I is $4.77\ \text{A}$ .

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3 years ago

A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio

Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

$d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at$

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.

Treat them separately.

At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.

Horizontal movement is not influenced by gravity.
acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

$d_x=10.0m\\d_y=3.0m$

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ? Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

$d_y=V_i_yt+\dfrac{1}{2}at^{2}$

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

$d_y=0+\dfrac{1}{2}at^{2}$

$d_y=\dfrac{1}{2}at^{2}$

From here we can derive the formula of time:

$t=\sqrt{\dfrac{2d_y}{a}}$

Now we just plug in what we know:

$t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s$

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

$vf_y=vi_y+at$

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

$0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s$

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

$d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt$

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em> $d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x$ </em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

$Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s$

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

$Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17$

The lion leapt at an angle of 50.16°.

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3 years ago

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Sophie [7]

Answer:

They sh0uld g0 t0 the reactor and then, see what the issue is...

Explanation:

then, see if they can fix the problem, im sorry if its wr0ng.

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