Answer:
the impulse experienced by the passenger is 630.47 kg
Explanation:
Given;
initial velocity of the car, u = 0
final velocity of the car, v = 9.41 m/s
time of motion of the car, t = 4.24 s
mass of the passenger in the car, m = 67 kg
The impulse experienced by the passenger is calculated as;
J = ΔP = mv - mu = m(v - u)
= 67(9.41 - 0)
= 67 x 9.41
= 630.47 kg
Therefore, the impulse experienced by the passenger is 630.47 kg
Answer:
B
Explanation:
this is because the neutrons do not have a charge, the things that have charge in an atom are electrons and protons.
and in the nucleus of an atom, there are protons and neutrons so you can see that A is not the answer
if you see the periodic table, you will know that the number of electrons and protons are equal, so the charges cancel each other out, hence the charge of an atom will be neutral
let me give you a tip which I got from my teacher, never write there is no charge in the atom, this suggests that there is no protons or electrons.
instead, write, the it is neutral
hope it helps if not please report it so that someone else gets to try it out
(a) Let
be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

where
is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

so that

Solve for
:

(b) The net force equation in part (a) leads us to the relation

so that
is directly proportional to the square root of
. As the radius
increases, the maximum linear speed
will also increase, so the cord is less likely to break if we keep up the same speed.
There's no such thing as "stationary in space". But if the distance
between the Earth and some stars is not changing, then (A) w<span>avelengths
measured here would match the actual wavelengths emitted from these
stars. </span><span>
</span><span>If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.
</span>In order to decide what's actually happening, and how that star is moving,
the trick is: How do we know the actual wavelengths the star emitted ?
Answer: I think it’s 20cm.