The angle theta of the spring is 31 degrees. To solve for this, show the equation which is equal to the 10lb ball. With this, the unknown will be the angle. Then transpose/transfer the terms in order to isolate the variable for the angle. First solve for s, then solve for angle theta. You will come up with s = .5849 ft and angle theta = 31.2629 or 31 degrees. Hope this helps.
Answer: <span>Protons do not move out of the nucleus of atoms although they repel each other.
Reasoning:
Remember that protons are particles with positive charge and they held together in the nucleus of the atom which is a tiny tiny region. As you know, like charges repel each other, which means that the protons exert a repulsion force.
That repulsion force tends to destabilize the atom. The only reason why the atom does not explode is because there is a there exists a force of attraction that exceeds the repulsion of the like charged protons. That is the strong nuclear force. Therefore, the strong nuclear forces are stronger than the electromagnetic force that tends to get the protons apart.
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The hierarchy of need was developed by Maslow. The pyramid is a motivational theory in psychology that argues that people is seeking to meet successively higher needs. The base of the pyramid is for the physiological needs to safety, belongingness, esteem, and finaly self-actualization.
Could you scroll up a little so I can see the first part of the question- I MIGHT be able to help you then. Thanks!
Define
g = 9.8 m/s², acceleration due to gravity, positive downward.
Assume that wind resistance may be neglected.
Frog A:
u = 0.551 m/s, launch velocity, upward.
When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.
If the time of flight is t, then
(0.551 m/s)*(t s) - 0.5*(9.8 m/s²)*(t s)² = 0
0.551t - 4.9t² = 0
t = 0, or t = 0.1124 s
t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.
Frog B:
Launch velocity is 1.75 m/s
When t = 0.1124 s, the position of the frog is
s = (1.75 m/s)(0.1124 s) - 0.5*(9.8 m/s²)*(0.1124 s)²
= 0.135 m
The velocity of frog B is
v = (1.75 m/s) - (9.8 m/s²)*(0.1124 s)
= 0.6485 m/s
Answer:
When frog A lands on the ground,
Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.
