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Leviafan [203]
2 years ago
15

Choose the incorrect statement about the proton: Group of answer choices The proton has the atomic mass of 1 amu The proton has

the same charge as the neutron. The proton has greater mass than an electron The proton and the neutron have approximately the same atomic mass
Physics
1 answer:
sergey [27]2 years ago
5 0

Answer:

The proton has the same charge as the neutron.

Explanation:

Because the proton has a positive charge whereas the neutron has no charge  at all.

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3 years ago
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Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the ci
Anettt [7]

Answer:

The radius is  r = 3.1905 \ m

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  d =  2.70 \ m

        The refractive index of the liquid is  n_i  =  1.31

Now the critical value is mathematically represented as

         \theta =  sin ^{-1} [\frac{1}{n_i} ]

substituting values

         \theta =  sin ^{-1} [\frac{1}{131} ]

         \theta  =  49.76^o

Using SOHCAHTOA rule we have that

         tan \theta =  \frac{ r}{d}

=>     r =  d * tan \theta

substituting values  

        r =  2.7 * tan (49.76)

        r = 3.1905 \ m

         

5 0
3 years ago
Please help <br>problems 2a.,2b.,3a.,and 3b.​
Darina [25.2K]

Answer:

2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]

Explanation:

2a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}

2b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]

3a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]

3b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}

8 0
3 years ago
Worth 15 points<br><br> Please answer which one matches for each conversion
FrozenT [24]

Answer:

distance - meters

speed - meters/seconds

time - seconds

velocity - meters/seconds

acceleration - meters/seconds²

5 0
2 years ago
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