Calculate the answer to the correct

Two protons, starting several meters apart, are aimed directly at each other with speeds of 2.00 * 105 m>s, measured relative

victus00 [196]

Explanation:

Since, the two protons are at the initial state of point a and point b. Hence, total mechanical energy at point a and point b is as follows.

$U_{a} + K_{a} = U_{b} + K_{b}$

or, $K_{a} = U_{b}$

where, $K_{a}$ = kinetic energy of two protons

So, $2(\frac{1}{2}mV^{2}_{a}) = \frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r_{b}}$

$r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}$

Putting the given values into the above formula we will calculate the value of $r_{b}$ as follows.

$r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}$

= $(9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}^{2} C}{1.67 \times 10^{-27} kg \times 2 \times 10^{5}}$

= $3.45 \times 10^{-12} m$

Now, we will calculate the maximum electric field as follows.

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r^{2}_{b}}$

= $9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}C)^{2}}{3.45 \times 10^{-12}}$

= $0.194 \times 10^{-4} N$

Therefore, we can conclude that the maximum electric force that these protons will exert on each other is $0.194 \times 10^{-4} N$ .

7 0

3 years ago

3. A certain horizontal east-west lined wire has a mass of 0.2kg per meter of length and carries a current I. Impressed on the w

erica [24]

Answer:

i = 4.9 A

Explanation:

The expression for the magnetic force in a wire carrying a current is

F = i L x B

bold letters indicate vectors.

The direction of the cable is towards the East, the direction of the magnetic field is towards the North, so the vector product is in the vertical direction (z-axis) upwards and the weight of the cable is vertical downwards. Let's apply the equilibrium condition

F - W = 0

i L B = m g

They indicate the linear density of the cable λ = 0.2 kg / m

λ = m / L

m = λ L

we substitute

i B = λ g

i = $\frac{ \lambda \ g}{B}$

let's calculate

i = 0.2 9.8 / 0.4

i = 4.9 A

3 0

2 years ago

What is the force exerted

Norma-Jean [14]

Answer:

I wish i saw the field shown to the right

Explanation:

3 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1

lesya [120]

Answer:

Incomplete question, check attachment for completed question

Explanation:

The force of attraction between two forces are given as

F=kQq/r²

4 0

3 years ago

What kind of potential energy allows you to throw a ball

Aleks [24]

Force.................

3 0

3 years ago

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