A testing instrument that's used to measure electrical signals in a circuit and display them as waveforms on a screen is called
1 answer:
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The image is missing (however it's not necessary to solve the problem).
The correct answer is A) decreases, because the gravitational force is inversely proportional to the square of the distance. In fact, the magnitude of the gravitational force between two object of mass M and m, at a distance d one from each other, is
where G is the gravitational constant. As can be seen from the formula, if the distance d between the two object increases, the intensity of the force decreases.
The correct answer is A) decreases, because the gravitational force is inversely proportional to the square of the distance. In fact, the magnitude of the gravitational force between two object of mass M and m, at a distance d one from each other, is
where G is the gravitational constant. As can be seen from the formula, if the distance d between the two object increases, the intensity of the force decreases.
Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w = =
= 18.38 rad/sec
Period of oscillation =
= 0.34 sec