The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.
Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).
At the beginning ... time=0 ... the position is 1 meter.
At the end ... time=5 ... the position is zero.
The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.
Answer:
Explanation:
Let's use the decay equation.
Where:
- A is the activity at t time
- A₀ is the initial activity
- λ is the decay constant
We know that 
So we have:
Therefore, the half-life of the source is 6 hours.
I hope it helps you!
(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
<h3>
Work done by the applied force</h3>
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
<h3>
Work done by the normal force</h3>
W = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
<h3>Work done force of gravity</h3>
The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
<h3> Work done by the net force on the block</h3>
∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
Learn more about work done here: brainly.com/question/8119756
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Answer:
6.75×10^13N
Explanation:
The electric force between the charges can be determined using coulombs law which states that 'the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of their distance between them.
Mathematically, F = kq1q2/r² where;
q1 and q2 are the charges
r is the distance between the charges
F is the force of attraction
k is the coulombs constant
Given q1 = 5C q2 = 15C r = 10cm = 0.1m k = 9×10^9Nm²/C²
Substituting the given values in the formula we have;
F =9×10^9×5×15/0.1²
F = 6.75×10^11/0.01
F = 6.75×10^13N
Therefore the electric force between them is 6.75×10^13N
