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Sloan [31]
3 years ago
8

A solid block of copper, which is a good conductor, has a cavity in its interior. Within the cavity, insulated from the conducto

r, is a point charge of 3.00 nCnC. The block of copper itself has an excess charge of -8.00 nCnC. How much charge is on the outer surface of the block and the surface of the cavity
Physics
1 answer:
Nikolay [14]3 years ago
4 0

Answer:

The charge on the outer surface of the block = -5.00 nC

The charge on the surface of the cavity (on the inner surface of the block) = -3.00 nC

Explanation:

The point charge within the cavity will induce a charge equal in magnitude and opposite in sign on the inside cavities of the copper block.

Charge of the point charge = 3.00 nC

Charge induced on the inner surface of the Copper block's cavity = -3.00 nC

Since the charge on a conductor should usually be neutral, the charge on the inner surface causes a charge equal in magnitude and also opposite in sign on the outer surface of the block; that is, 3.00 nC.

But this block already has an excess charge of -8.00 nC (which resides on the surface because excess charge for conductors reside on the surface of the conductors)

So, net charge on the outer surface of the Copper block = -8.00 + 3.00 = -5.00 nC.

Hope this Helps!!!

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A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−
Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

8 0
3 years ago
Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f
RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

The force exerted on each piston is just equal to the weight of the corresponding mass: F=W=mg, where m is the mass and g is the gravitational acceleration. So the equation becomes

\frac{m_A g}{A_A}=\frac{m_B g}{A_B}

Now we can rewrite the mass as the product of volume, V, times density, d:

\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}

We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

So we can further re-arrange the equation (and simplify g as well):

\frac{V_A d_A}{1}=\frac{V_B d_B}{2}

\frac{d_A}{d_B}=\frac{V_B}{2V_A}

We are also told that block B has bigger volume than block A: V_B > V_A. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0
3 years ago
Question 6. B) and c)
maks197457 [2]
The formula for both is v(t) = v0 + a*t

b) v(8) = 0 + 6m/s^2 *8s = 48 m/s

now we know the beginning (2) and end speed (14), but not the time:

c) 14 = 2 + 1.5*t => t = (14-2)/1.5 = 8 seconds
4 0
4 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.
love history [14]

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

3 0
4 years ago
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