Answer:
x = A sin ω t describes the displacement of the particle
v = A ω cos ω t
a = -A ω^2 sin ω t
a (max) = -A ω^2 is the max acceleration (- can be ignored here)
ω = (K/ m)^1/2 for SHM
F = - K x^2 restoring force of spring
K = 4.34 / .0745^2 = 782 N / m
ω = (782 / .297)^1/2 = 51.3 / sec
a (max) = .0745 * 782 / .297 = 196 m / s^2
The electric force between two charged particles can be increased by decreasing the distance between the two particles.
<h3>How to increase electric force between two charged particles.</h3>
The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while
increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.
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Answer:
A. 28.42 m/s
B. 41.21 m.
Explanation:
A. Determination of the initial velocity of the ball:
Time (t) to reach the maximum height = 2.9 s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) =?
Thus, we can obtain the initial velocity of the ball as follow:
v = u + gt
0 = u + (–9.8 × 2.9)
0 = u – 28.42
Collect like terms
u = 0 + 28.42
u = 28.42 m/s
Therefore, the initial velocity of the ball is 28.42 m/s.
B. Determination of the maximum height reached.
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) = 28.42 m/s.
Maximum height (h) =?
Thus, we can obtain the maximum height reached by the ball as follow:
v² = u² + 2gh
0² = 28.42² + (2 × –9.8 × h)
0 = 807.6964 + (–19.6h)
0 = 807.6964 – 19.6h
Collect like terms
0 – 807.6964 = – 19.6h
– 807.6964 = – 19.6h
Divide both side by – 19.6
h = –807.6964 / –19.6
h = 41.21 m
Therefore, the maximum height reached by the ball is 41.21 m
Answer:
<h3>The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST</h3>
Explanation:
Momentum = mass*velocity of a body
For a 4.00 kg ball is moving at 4.00 m/s to the EAST, its momentum = 4*4 = 16kgm/s
For a 6.00 kg ball is moving at 3.00 m/s to the NORTH;
its momentum = 6*3 = 18kgm/s
Total momentum = The resultant of both momentum
Total momentum = √16²+18²
Total momentum = √580
total momentum = 24.1kgm/s
For the direction:
![\theta = tan^{-1} \frac{y}{x}\\\theta = tan^{-1} \frac{18}{16}\\ \theta = tan^{-1} 1.125\\\theta = 48.4^{0}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5Cfrac%7By%7D%7Bx%7D%5C%5C%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5Cfrac%7B18%7D%7B16%7D%5C%5C%20%5Ctheta%20%3D%20%20tan%5E%7B-1%7D%201.125%5C%5C%5Ctheta%20%3D%2048.4%5E%7B0%7D)
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST