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katrin2010 [14]
3 years ago
5

A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi

th negative charge −e distributed uniformly around the ring (representing the electron in orbit around the proton). Find the magnitude of the total electric field due to this charge distribution at a point a distance a from the proton and perpendicular to the plane of the ring.
Physics
1 answer:
Norma-Jean [14]3 years ago
8 0

Answer:

Now e is due to the ring at a

So

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

So

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

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If the octupus had twelve arms then how many arms would the crab have show your work
Alexandra [31]

Answer:

I believe  that the crab would have 14 arms if the octopus had 12 arms

Explanation:

4 0
2 years ago
Two point charges, with charge magnitudes q and ????, are placed a distance r apart. In this arrangement, each point charge expe
sammy [17]

Answer:

1)  Q ’= 8 Q ,  2)    q ’= 16 q ,  3)   r ’= ¾ r

Explanation:

For this exercise we will use Coulomb's law

      F = k q Q / r²

It asks us to calculate the change of any of the parameters so that the force is always F

Original values

                q, Q, r

Scenario 1

      q ’= 2q

       r ’= 4r

     F = k q ’Q’ / r’²

we substitute

     F = k 2q Q ’/ (4r)²

     F = k 2q Q '/ 16r²

we substitute the value of F

      k q Q / r² = k q Q '/ 8r²

       Q ’= 8 Q

Scenario 2

       Q ’= Q

       r ’= 4r

we substitute

      F = k q ’Q / 16r²

      k q Q / r² = k q’ Q / 16 r²

      q ’= 16 q

Scenario 3

      q ’= 3/2 q

      Q ’= ⅜ Q

we substitute

        k q Q r² = k (3/2 q) (⅜ Q) / r’²

        r’² = 9/16 r²

        r ’= ¾ r

6 0
3 years ago
Newtons third law says that if Robert exerts a _______ of 1000 Newtons on an object, it will exert an equal and opposite _______
Free_Kalibri [48]

Answer: force, force

Explanation:

Newton’s third law states that there is an equal and opposite force

I took the test too

7 0
3 years ago
Why do you not come to thermal equilibrium on a cold day
Tju [1.3M]

Two physical systems are in thermal equilibrium if no heat flows between them when they are connected by a path permeable to heat. Thermal equilibrium obeys the zeroth law of thermodynamics. A system is said to be in thermal equilibrium with itself if the temperature within the system is spatially and temporally uniform.

Systems in thermodynamic equilibrium are always in thermal equilibrium, but the converse is not always true. If the connection between the systems allows transfer of energy as heat but does not allow transfer of matter or transfer of energy as work, the two systems may reach thermal equilibrium without reaching thermodynamic equilibrium.

3 0
3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0
3 years ago
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