Answer:
the charge of the particle is 2.47 x 10⁻¹⁹ C
Explanation:
Given;
mass of the particle, m = 6.64 x 10⁻²⁷ kg
velocity of the particle, v = 8.7 x 10⁵ m/s
strength of the magnetic field, B = 1.3 T
radius of the circle, r = 18 mm = 1.8 x 10⁻³ m
The magnetic force experienced by the charge is calculated as;
F = ma = qvB
where;
q is the charge of the particle
a is the acceleration of the charge in the circular path
Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C
Ah ha ! Very interesting question.
Thought-provoking, even.
You have something that weighs 1 Newton, and you want to know
the situation in which the object would have the greatest mass.
Weight = (mass) x (local gravity)
Mass = (weight) / (local gravity)
Mass = (1 Newton) / (local gravity)
"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest. This is the
clue that gives it away.
If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:
"Fine ! Great ! Golly gee, that's sure generous of you.
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there. And if you don't mind, be quick about it."
The local acceleration of gravity on Pluto is 0.62 m/s² ,
but on Earth, it's 9.81 m/s.
So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth.
That's almost 3.6 pounds of gold, worth over $57,000 !
It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid. Wherever he's willing to go
that has the smallest gravity. That's the place where the largest
mass weighs 1 Newton.
Answer:
The answer is B: to Amplify the sound
Here we want to study how the linear charge density changes as we change the measures of our body.
We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.
<em>I will take two assumptions:</em>
The charge is homogeneous, so the density is constant all along the wire.
As we work with a linear charge density we work in one dimension, so the wire "has no radius"
Originally, the wire has a charge Q and a length L.
The linear charge density will be given by:
λ = Q/L
Now the length of the wire is stretched to 10 times the original length, so we have:
L' = 10*L
We want to find the value of Q' such that λ' (the <u>linear density of the stretched wire</u>) is still equal to λ.
Then we will have:
λ' = Q'/L' = Q'/(10*L) = λ = Q/L
Q'/(10*L) = Q/L
Q'/10 = Q
Q' = 10*Q
So the new <u>charge must be 10 times the original charge</u>, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.
If you want to learn more, you can read:
brainly.com/question/14514975
.The path of a celestial body or an artificial satellite as it revolves around another body due to their mutual gravitational <span>attraction.</span>
