Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts
Newton’s fifth law says so i’m sorry it’s just logic
Answer:
50 Hz is the answer because 60 m/s divided by 1.2 meters is 50.
Hope that helps you!!! Please give me Brainliest!!!!
Answer:
If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.
Explanation:
By momentum conservation, <em>and given the bullit and the recoil are in a straight line</em>, the momentum analysis will be <em>unidimentional</em>. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so

now we clear
and use the given data to get that

<em>But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s</em>, then <u>if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained</u>, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity
too.
Finally, using
we calculate the kinetic energy as

X-rays have high energy and can penetrate matter that light cannot.