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Anastasy [175]
3 years ago
7

Determine the electrical force that two protons in the nucleus of a helium atom exert on each other when separated by 2.8×10−15m

. Express your answer with the appropriate units.
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
8 0
<h2>Answer:</h2>

29.4N

<h2>Explanation:</h2>

By Coulomb's law, the electrical force of attraction/repulsion, F,  between two charges of magnitudes Q₁ and Q₂, is directly proportional to the product of the magnitudes of these charges and inversely proportional to the square of the distance, r, between these charges. i.e;

F ∝ Q₁ x Q₂ / r²

F = k x Q₁ x Q₂ / r²    -------------------(i)

Where;

k = proportionality constant = 8.99 x 10⁹Nm²/C²

<em>From the question;</em>

Two protons are given;

Let the charge of first proton be Q₁

Let the charge of second proton be Q₂

The charge of a proton  = 1.602 x 10⁻¹⁹C

=> Q₁ = Q₂ =  1.602 x 10⁻¹⁹C

<em>Also;</em>

r = 2.8 x 10⁻¹⁵m

<em>Substitute these values into equation (i) s follows;</em>

F = 8.99 x 10⁹ x 1.602 x 10⁻¹⁹ x  1.602 x 10⁻¹⁹ / (2.8 x 10⁻¹⁵)²

F = 8.99 x 10⁹ x 1.602 x 10⁻¹⁹ x  1.602 x 10⁻¹⁹ / (7.84 x 10⁻³⁰)

F = 23.07 x 10⁻²⁹ / (7.84 x 10⁻³⁰)

F = 29.4N

Therefore, the electrical force that the two protons exert on each other is 29.4N

lisabon 2012 [21]3 years ago
3 0

Answer:

6.27 x 10^10 N  

Explanation:

Since the mass of the proton is 1.602 x 10-19 C, therefore the electric force between two protons separated by a distance:  

r = 2.8×10^−15m  

is given as following:  

F=k*║q_1║*║q_2║/r^2

  =k*║q║*║q║/r^2

  =k*║q║^2/r^2

  =k*q^2/r^2

  = 8.99 x 10^9 Nm^2c^2*[(1.67 x 10^-19 c)^2/ (2.8 x 10^-15 m)^2  

  = 6.27 x 10^10 N  

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6 0
2 years ago
What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a
Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

A = W + \dfrac{W}{2}

A =\dfrac{3W}{2}

A =\dfrac{3mg}{2}

When the car is at the bottom,  the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

N = m g + \dfrac{mv^2}{r}

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v = \sqrt{\dfrac{rg}{2}}

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8 0
2 years ago
In physics what does 7.56 × 5.746 equal ?
lapo4ka [179]

Answer:

43.43

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Ans: 43.43.

7 0
3 years ago
A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

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Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
3 years ago
A 1400-kg car is moving at a speed of 25m/s. how much kinetic energy does the car have?
Luba_88 [7]

Answer:

437.5Kjoules

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K.E=half multiply by mass multiply by square of velocity

=437.5Kjoules

3 0
2 years ago
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