Answer:
Atomic name is your answer.
Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.
The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
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Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>
<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>
<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>
so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
Answer:
The the analysis for the free fall part should be done under the constant acceleration.
Explanation:
In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.
Therefore, the the analysis for the free fall part should be done under the constant acceleration.
The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.
Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get
