The solution you should use is Hooke's law: F=-kx
It should have the same signs because they repel due to the stretch of the spring.
a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be
<span>F = kx
270 N/m x 0.38 m = 102.6 N
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b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force.
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.
3
The amplitude of oscillations has no effect on their time period.
First one is B, second one is A
Answer:
The answer is Gravitational pull
