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3 years ago

A sphere has a surface uniformly charged with 3.30 C. At what distance from its center is the potential 4.50 MV

Physics

1 answer:

Bogdan [553]3 years ago

6 0

Answer:

r = 6.6 x 10³ m = 6600 m

Explanation:

The potential at a distance from a charged sphere can be given as follows:

$V = \frac{kq}{r}\\\\r = \frac{kq}{V}$

where,

r = distance = ?

k = Colomb Constant = 9 x 10⁹ Nm²/C²

q = charge on sphere = 3.3 C

V = potential = 4.5 MV = 4.5 x 10⁶ V

Therefore,

$r = \frac{(9\ x\ 10^9\ Nm^2/C^2)(3.3\ C)}{4.5\ x\ 10^6\ V}$

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dangina [55]

Answer:

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3 years ago

Read 2 more answers

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v

Semenov [28]

Answer:

A)Object 1 has the greater magnitude of its momentum.

B)The objects 2 have the greater kinetic energy.

Explanation:

For object 1 :

v₁ = v ,m₁ = 2 m

For object 2 :

$v_2=2\sqrt{v}$ ,m₂=m

We know that linear momentum given as

P = M V

M=Mass , V=Velocity

For object 1 :

P₁ =m₁ v₁

P₁ =2 m v

For object 2

$P_2=m_2v_2$

$P_2=2m\sqrt {v}$

We can say that object 1 have more momentum.

The kinetic energy

$KE_1=\dfrac{1}{2}m_1v_1^2$

$KE_1=\dfrac{1}{2}\times 2m\times v^2$

$KE_1=mv^2$

$KE_2=\dfrac{1}{2}m_2v_2$

$KE_2=\dfrac{1}{2}\times m\times 4v^2$

$KE_2=2mv^2$

Therefore both the object 2 have higher kinetic energy.

7 0

3 years ago

A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux throu

ioda

Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

We assume that the magnetic field in the solenoid is constant.

The magnetic flux is given as:

$\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb$