Answer:
The work done required on the coin during the displacement is 21.75 w.
Explanation:
Given that,
A coin slides over a friction-less plane i.e friction force = 0.
The co-ordinate of the given point is (1.40 m, 7.20 m).
The position vector of the given point is represented by 
.
The displacement of the coin is
The force has magnitude 4.50 N and its makes an angle 128° with positive x axis.
Then x component of the force = 4.50 cos128°
The y component of the force = 4.50 sin128°
Then the position vector of the force is
We know that,
work done is a scalar product of force and displacement.
=(-2.77×1.40+ 3.56×7.20) w
=21.75 w
The work done required on the coin during the displacement is 21.75 w.
<span>(1KM x 1000M/KM)/ 850 M/s = 1.18 seconds</span>
That's right, the correct answer is
<span>A) The isotopes have a long half-life and only remain radioactive for a long time period
The half life of an isotope is the time it takes for the amount of the sample to reduce to half of its initial value. If an isotope has a long half-life, it means it takes a long time to reduce down to a significant level, so it will remain radioactive for a long time period.</span>
Answer:
20.25 m
Explanation:
- <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.
That is;
<em><u>ac = v²/r</u></em>
<em> </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m</u></em>
Therefore;
r = v²/ac
= 27²/36
= 20.25 m
Hence the radius is 20.25 meters
