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weqwewe [10]
3 years ago
10

A hollow, conducting sphere with an outer radius of 0.240 m and an inner radius of 0.200 m has a uniform surface charge density

of +6.47 × 10⁻⁶ C/m². A charge of -0.900 μC is now introduced into the cavity inside the sphere.
A. What is the new charge density σ on the outside of the sphere? Express your answer with the appropriate units.
B. Calculate the strength of the electric field E just outside the sphere. Express your answer with the appropriate units.
C. What is the electric flux through a spherical surface just inside the inner surface of the sphere? Express your answer with the appropriate units.
Physics
1 answer:
GarryVolchara [31]3 years ago
5 0
A. What is the new charge density
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Brilliant_brown [7]

Answer:

Explanation:

x

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3 years ago
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the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
3 years ago
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

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4 0
3 years ago
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An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the
Eduardwww [97]

Answer: 1.14 N

Explanation :

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In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.

We can get the gravity force as follows:

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The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:

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Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N

Equating both sides of Newton´s 2nd Law in the vertical direction:

T + Fg = Fb

T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N

6 0
3 years ago
Please help with this!!!!!
34kurt
(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K)  I hoped that helped
4 0
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