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evablogger [386]
2 years ago
5

A solid nonconducting sphere of radiusRcarries a chargeQdistributed uniformly throughout itsvolume. At a certain distancer1(r1&l

t; R) from the center of the sphere, the electric field has magnitudeE.If the same chargeQwere distributed uniformly throughout a sphere of radius 2R, the magnitude of theelectric field at the same distancer1from the center would be equal to______
Physics
1 answer:
lara [203]2 years ago
3 0

Answer:

E' = \frac{1}{8} E

Explanation:

Given data:

first case

Distance of electric field from center of sphere is r_1 <R

Electric field at r_1< R

E = \frac{kQr_1}{R^3}

second case

Distance of electric field from centre of sphere is r_1 < 2R

Electric field at r_1< 2R

E' = \frac{kQr_1}{8R^3}

so, we have

E' = \frac{1}{8} E

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Which statement best describes perigee?
pantera1 [17]

Answer:

A. The closest point in the Moon's orbit to Earth

Explanation:

The perigee is defined as the closest point in the orbit of an object (such as a satellite) from the centre of the Earth. In this case, the Earth's satellite is the Moon, so the perigee is defined as the closest point in the Moon's orbit to Earth. so option A is the correct one.

Let's see instead the names of the other options:

B. The farthest point in the Moon's orbit to Earth  --> this point is called apogee

C. The closest point in Earth's orbit of the Sun  --> this point is called perihelion

D. The Sun's orbit that is closest to the Moon --> this point has no specific name

8 0
3 years ago
Read 2 more answers
On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around
Mekhanik [1.2K]

Answer:

v\approx 16.956\,\frac{m}{s}

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

\Sigma F = f = m\cdot \frac{v^{2}}{R}

The maximum speed is:

v = \sqrt{\frac{f\cdot R}{m} }

v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }

v\approx 16.956\,\frac{m}{s}

7 0
3 years ago
As a stands, her entire weight is momentarily placed on the heels of her high-heeled shoes. Calculate the pressure exerted on th
Lena [83]

Answer:

Pressure will be 6072449.952Pa

Explanation:

We have given mass of the women m = 65 kg

Radius of the heels r = 0.578 cm = 0.00578 m

We have to find the pressure

We know that pressure is given by

P=\frac{F}{A}=\frac{mg}{A}

So force F = mg = 65×9.8 = 637 N

Area A=\pi r^2=3.14\times 0.00578^2=1.049\times 10^{-4}m^2

So pressure p=\frac{637}{1.049\times 10^{-4}}=6072449.952Pa

5 0
3 years ago
A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in
Leno4ka [110]

Answer:

The  electric field  is 35\cos(90\pi t)\ mV/m

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length n= 1.65\times10^{3}

Current I = 6.00\sin 90\pi t

We need to calculate the induced emf

\epsilon =\mu_{0}nA\dfrac{dI}{dt}

Where, n = number of turns per unit length

A = area of cross section

\dfrac{dI}{dt}=rate of current

Formula of electric field is defined as,

E=\dfrac{\epsilon}{2\pi r}

Where, r = radius

Put the value of emf in equation (I)

E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}....(II)

We need to calculate the rate of current

I=6.00\sin 90\pi t....(III)

On differentiating equation (III)

\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)

Now, put the value of rate of current in equation (II)

E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}

E=35\cos(90\pi t)\ mV/m

Hence, The  electric field  is 35\cos(90\pi t)\ mV/m

7 0
2 years ago
Read 2 more answers
21. (a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500-μC charge and flies due west at a
beks73 [17]

Answer:

Explanation:

The picture attached shows the full explanation and i hope it helps. Thank you

6 0
3 years ago
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