Answer:
Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?
Search instead for Two point charge q, seperated by 1.5cm have change value of +2.0 and -4.0N/C respectively what is the magnitude of the Electric force midway between them?
Answer: Due that we don't know the initial speed after hitting the ball, we are going to accept that the ball goes up for half of the time and then falls during other half part, that is 3.0 seconds each. Then we know that ball's movement is ruled by the acceleration of gravity formula, as follows: H = Vi * T + 1/2 * g * T^2 V = Vi + g * T where: H is height, Vi initial speed, g gravity acceleration and T time When we only consider the second half of the trajectory, we have that initial speed at the top of that movement is zero, because ball goes up till top, where stops and starts to go down, so : H = 0 * 3 + 1/2 * 32 * 3^2 = 144 ft. So the height of the pop-up is 144 feet.
Answer:
a) μ = 0.475
, b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - 
= 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433
