If Teresa's daughter is my daughters mother who am I to Teresa?

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

4 years ago

PLEASE HELP ME WITH THSES QUESTIONS 3 SENTENCES PLEASE I BEG YOU!

JulsSmile [24]

Question 2 is because the passengers have inertia, which is a tendency to resist change in motion

6 0

3 years ago

What is conduction and induction?

Romashka [77]

Conduction involves physical contact to charge, well induction does not.
Learn more at: <span>www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Conduction</span>

8 0

3 years ago

Read 2 more answers

The size of a balloon increases when the pressure inside it increases. The balloon gets bigger when it is left in the heat from

Masteriza [31]

Explanation:

This happens because the gas inside tend to expand because its temperature gets higher.

This is why the balloon that is put in a freezer for too long tend to gets smaller, because the gas temperature that is inside the balloon decreases.

(you can try it at home)

It is related to the temperature of the gas.

5 0

3 years ago

A ball is in free fall after being dropped. What willthe speed of the ball be after 2 seconds of free fall?

Mazyrski [523]

So, the speed of the ball after 2 seconds after free fall is <u>20 m/s</u>.

<h3>Introduction</h3>

Hi ! I'm Deva from Brainly Indonesia. In this material, we can call this event "Free Fall Motion". There are two conditions for free fall motion, namely falling (from top to bottom) and free (without initial velocity). Because the question only asks for the final velocity of the ball, in fact, we may use the formula for the relationship between acceleration and change in velocity and time. In general, this relationship can be expressed in the following equation :

$\boxed{\sf{\bold{a = \frac{v_2 - v_1}{t}}}}$

With the following conditions :

a = acceleration (m/s²)
$\sf{v_2}$ = speed after some time (m/s)
$\sf{v_1}$ = initial speed (m/s)
t = interval of time (s)

<h3>Problem Solving</h3>

We know that :

a = acceleration = 9,8 m/s² >> because the acceleration of a falling object is following the acceleration of gravity (g).
$\sf{v_1}$ = initial speed = 0 m/s >> the keyword is free fall
t = interval of time = 2 s

What was asked :

$\sf{v_2}$ = speed after some time = ... m/s

Step by step :

$\sf{a = \frac{v_2 - v_1}{t}}$

$\sf{(a \times t) + v_1 = v_2}$

$\sf{(10 \times 2) + 0 = v_2}$

$\boxed{\sf{v_2 = 20 \: m/s}}$

So, the speed of the ball after 2 seconds after free fall is 20 m/s.

8 0

2 years ago