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NeX [460]
3 years ago
13

For a charged hollow metal sphere with total charge Q and radius R centered on the origin: (Determine whether each of the follow

ing statements is correct or incorrect.) The field inside the shell is zero. The field for r > R will be the same as the field of a point charge, Q, at the origin The charge is on the inside surface. Only + charges can be on the outside surface. The E-field on the outside is perpendicular to the surface. Inside the metal the field is strongest.
Physics
1 answer:
Sloan [31]3 years ago
5 0

Answer:

1. The field inside the shell is 0; TRUE

2. The field for r>R will be the same as a field for the point charge Q, at the origin: TRUE

3. only positive charges can be on the outside surface: FALSE (Negative can too!!!)

4. The field on the outside is perpendicular to the surface: TRUE

Explanation:

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5 0
3 years ago
A 6 V battery is connected to a 24 ohm resistor to create a circuit. The 6 V battery is then replaced with a 12 V battery. How d
11Alexandr11 [23.1K]

Answer:

1.) The current is doubled

2.) moving a magnet up and down near the wire

3.) An electric current in the wire produces a magnetic field.

4.) Distance between Particles (m)

Explanation:

1.) When 6 V battery is connected to a 24 ohm resistor to create a circuit, using ohms law, V = IR

Current I = 6/24 = 0.25 A

When the The 6 V battery is replaced with a 12 V battery,

Current I = 12/24 = 0.5 A

Therefore, The current is doubled

2.) Electric current will be induced when moving a magnet up and down near the wire

3.) An electric current in the wire produces a magnetic field.

4.) Distance between Particles (m)

The force of attraction between two different masses is inversely proportional to the square of the distance between them.

6 0
4 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

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4 0
3 years ago
19. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to
Aleks [24]

Recall this gas law:

\frac{P₁V₁}{T₁} = \frac{P₂V}{T₂}

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V₁ and V₂ are the initial and final volumes.

T₁ and T₂ are the initial and final temperatures.

Given values:

P₁ = 475kPa

V₁ = 4m³, V₂ = 6.5m³

T₁ = 290K, T₂ = 277K

Substitute the terms in the equation with the given values and solve for Pf:

\frac{475*4}{290} = \frac{P₂*6.5}{277}

<h3>P₂ = 279.2kPa</h3>
8 0
3 years ago
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