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
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Answer:
1/2 M V^2 = .1 M g H where 10% of PE goes into KE
V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2
V = 31.1 m/s increase in speed during descent
1 km / hr = 1000 m / 3600 sec = .278 m/s
V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr
Potential energy is highest when the car is released at the top of the ramp. The correct answer is option C
Potential energy is the energy possessed by a body when the body is at rest. Potential energy is at time called gravitational potential energy which as a product of mass of the body, acceleration due to gravity and the height attained by the body. That is,
P.E = mgh
When a car is moving down a ramp, the potential energy of the car can never remain the same except the car stop at a certain point.
Whenever a car is moving down a ramp, the potential energy of the car will be highest when the car is release at the top of the ramp. And lowest when the car reaches the bottom of the ramp.
The statement that is correct about the potential energy of a car moving down a ramp is:
Potential energy is highest when the car is released at the top of the ramp.
Therefore, the correct answer is option C
Learn more here: brainly.com/question/17400615
you are correct it is B those all play a key factor in motion and direction
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