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kotykmax [81]
3 years ago
14

A proton starts from rest and gains 8.35 x 10^-14 joule of kinetic energy as it accelerates between points A and B in an electri

c field. The potential difference between points A and B in the electric field is approximately

Physics
2 answers:
Schach [20]3 years ago
8 0

Answer : The potential difference between points A and B in the electric field is, 5.22\times 10^{5}V

Explanation :

The relation between kinetic energy and potential difference is:

K.E=q\times V

where,

K.E = kinetic energy  = 8.35\times 10^{-14}J

q = charge on proton = 1.602\times 10^{-19}C

V = potential difference = ?

Now put all the given values in the above formula, we get:

8.35\times 10^{-14}J=1.602\times 10^{-19}C\times V

V=5.22\times 10^{5}V

Therefore, the potential difference between points A and B in the electric field is, 5.22\times 10^{5}V

antoniya [11.8K]3 years ago
6 0

Answer:

5.22 x 10^5 V

Explanation:

guessed on castle learning and got it right

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In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
3 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
A device for exercising the upper leg muscle is shown above, together with a schematic representation of an equivalent lever sys
yuradex [85]

yo mamaaaaaaaaaaaaaaa

4 0
3 years ago
How do atoms of copper differ from atoms of aluminum? *
Degger [83]

Answer:

The copper atoms are heavier than the aluminum atoms. The copper atoms are smaller than the aluminum atoms so more copper atoms fit in the same volume. 2. Copper is more dense than aluminum.

HOPE THIS HELPS !!!!!

8 0
3 years ago
Please help asap!!!!!!!!!!!
Tomtit [17]
The answer would be 20000

The answer in standard form/scientific notation would be 2 x 10^4

(The exponent is 4 because that's how many digits after the 2 there is)
8 0
2 years ago
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