Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi
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Answer:
The maximum height of the arrow is 42 (and the units given for the height)
Explanation:
Everything is easier if you make a graph, you can give values to t and replace that values in the function, for example:
When t=0
h(0)=26
If you give some values to t you can see how the trajectory of the arrow is (please look the graphic below)
Now, to find the maximum you have to find the derivative of the function that describes the height of the arrow:
Then you have to take the derivative, and equals to zero to find t:
-32t+32=0
-32t=32
t=1
That is in the time of 1 second the arrow has its maximum height.
Now you have to replace this value in the original function, to find the height of the arrow:
h(1)=-16+32+36
h(1)=42
Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;
solving this two equations together;
where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t
Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.