Answer:
m = 63 grams
Explanation:
ω = 10 cycles/s(2π radians/cycle) = 20π rad/s
ω = √(k/m)
m = k/ω² = 250/(20π)² = 0.06332... kg
Answer is B- 200 m
Given:
m (mass of the car) = 2000 Kg
F = -2000 N
u(initial velocity)= 20 m/s.
v(final velocity)= 0.
Now we know that
<u>F= ma</u>
Where F is the force exerted on the object
m is the mass of the object
a is the acceleration of the object
Substituting the given values
-2000 = 2000 × a
a = -1 m/s∧2
Consider the equation
<u>v=u +at</u>
where v is the initial velocity
u is the initial velocity
a is the acceleration
t is the time
0= 20 -t
t=20 secs
s = ut +1/2(at∧2)
where s is the displacement of the object
u is the initial velocity
t is the time
v is the final velocity
a is the acceleration
s= 20 ×20 +(-1×20×20)/2
<u>s= 200 m</u>
My personal understanding and opinion is that ALL of those questions
should be part of an assessment of Physical Activity Readiness.
It is very difficult for an atom to accept a proton. It can only be done under very special circumstances. So A and C are both incorrect. I don't see how D is possible. The atom does lose 1 electron, but how it gets 21 is think air.
The answer is B which is exactly what happens.
Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
- When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
- While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
- And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.
