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gladu [14]
3 years ago
15

It was a children versus grown-ups competition at school. One event required the adult to throw a basketball as far as he could.

When it was the child's turn, he had to throw a baseball instead. The event coordinator said, "It's only fair!" Based on Newton's laws, was this event fair?
(A)Yes, because an adult can apply more force than a child can. Making the adult's object bigger than the child's object helps to even out the acceleration.
(B)No, because a basketball is bigger than a baseball, and objects that are bigger accelerate slower.
(C)Yes, because acceleration doesn't depend on the size of the object.
(D)No, because you must have the same size object in a competition.
Physics
2 answers:
Rudik [331]3 years ago
7 0

the answer is: (D) No, because you must have the same size object in a competition.

Cerrena [4.2K]3 years ago
3 0
D Because If Your Going To Have A Contest Its Ganna Have To Be The Same Objectives For Both Contenders 
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Lesechka [4]

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 2×10​⁵​​ 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;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

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

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

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