Answer:
good question, did you press any certain buttons?
Explanation:
so it has a balanced center of gravity
hope fully this helps
Answer:
the nucleus of the atom you have conserved almost all its mass in a very small volume
Explanation:
In Rutherforf's experiment, most alpha particles had few deflections, eg a few of them were deflected at small angles, a much smaller number deflected at large angles, and very few were deflected.
The explanation that I found for this is that if all the mass and the positive charge of the atom is in a small volume, they could specifically calculate this the relationship of the volumes,
V_atom / V_nucleo = 10⁴
Therefore, the phrase that explains the observing phenomenon is that the nucleus of the atom has concentrated almost all its mass, very small emolument and in that same volume is all the positive charge of the atoms.
In summary, the phrase that describes the process is:
the nucleus of the atom you have conserved almost all its mass in a very small volume
According to Newton's second law of motion, acceleration of the object is directly proportional to the net force in the object. It is expressed as:
F = ma
where F is the net force, m is the mass of the object and a is the acceleration.
20N = 5/1000 kg (a)
a = 4000 m/s²
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
