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anygoal [31]
3 years ago
12

Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the

protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart. What is the magnitude of the electric force between two protons separated by 2.00 ✕ 10-15 m

Physics
2 answers:
SVETLANKA909090 [29]3 years ago
6 0

The magnitude of the electric force between two protons separated by 2.00 × 10⁻¹⁵ m is about 57.6 Newton

\texttt{ }

<h3>Further explanation</h3>

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

\texttt{ }

There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }

<em>F = electric force (N)</em>

<em>k = electric constant (N m² / C²)</em>

<em>q = electric charge (C)</em>

<em>r = distance between charges (m)</em>

<em>The value of k in a vacuum = 9 x 10⁹ (N m² / C²)</em>

Let's tackle the problem now !

\texttt{ }

<u>Given:</u>

distance between protons = d = 2 × 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

<u>Unknown:</u>

electric force = F = ?

<u>Solution:</u>

F = k \frac{q_1 q_2}{(d)^2}

F = k \frac{q^2}{(d)^2}

F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(2 \times 10^{-15})^2}

F = 57.6 \texttt{ Newton}

\texttt{ }

<h3>Learn more</h3>
  • The three resistors : brainly.com/question/9503202
  • A series circuit : brainly.com/question/1518810
  • Compare and contrast a series and parallel circuit : brainly.com/question/539204

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Static Electricity

castortr0y [4]3 years ago
4 0

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}.

Now, making the substitutions

d \ = \ 2.00 * 10 ^{-15} \ m,

q_1 = q_2 = 1.60 * 10 ^ {-19} \ C,

\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2},

we can find:

F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}.

F \ = 57.536 N.

Not so big for everyday life, but enormous for subatomic particles.

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Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed
sergij07 [2.7K]

Answer:

v_{B0}=15.73 m/s

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

p_{0x}=p_{fx}

m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}  

m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40) (1)

y-coordinate

p_{0y}=p_{fy}

m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}  

m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40) (2)

We can divide equations (2) and (1):

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)

v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)

v_{B0}=\frac{1200*25}{1600}*tan(40)

v_{B0}=15.73 m/s

I hope it helps you!

           

4 0
3 years ago
Read 2 more answers
A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has
Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0
3 years ago
Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

1.28 sine r = 0.4067

Sine r = 0.4067/1.28

r = sine^-1(0.317)

r = 18.481

= 18.5⁰

4 0
2 years ago
A satellite of Mars, called Phobos, has an orbital radius of 9.4 ✕ 106 m and a period of 2.8 ✕ 104 s. Assuming the orbit is circ
FromTheMoon [43]

Answer:

6.27*10^{23}kg

Explanation:

assume

M= mass of Mars

m=mass of phobos

r=orbital radius

T=period

we can apply F=ma to this orbital motion (considering the cricular motion laws)

where,

F=\frac{GMm}{r^{2} }  and a=rω^2

where ω=\frac{2\pi }{T} and G is the universal gravitational constant.

G = 6.67 x 10-11 N m2 / kg2

F=ma\\\frac{GMm}{r^{2} }=mr(\frac{2\pi }{T} )^{2}\\  M=\frac{r^{3}}{G}  (\frac{2\pi }{T} )^{2}\\M=\frac{(9.4*10x^{6} )^{3}*(2\pi )^{2} }{(2.8*10^{4}) ^{2} *6.67*10^{-11} } \\M=6.27*10^{23}kg

6 0
3 years ago
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madreJ [45]
We will apply the Newton's second Law so the we will be able to find the acceleration.
F (tot) = ma
a = F(tot) /  m
a = 32.0 N / 65.0 kg = 0.492 m/s^2
Approximately 0.492 m/s^2 is her initial acceleration if she is initially stationary and wearing steel-bladed skates.

7 0
3 years ago
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