Answer:
Explanation:
force constant of spring k = force / extension
= 35.6 / 0.5
k = 71.2 N / m
angular frequency ω of oscillation by spring mass system
where m is mass of the body attached with spring
Putting the values
ω = 3.77 radian / s
The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2
so the equation for displacement from equilibrium position that is middle point can be given as follows
x = .5 sin ( ω t + π / 2 )
= 0.5 cos ω t
= 0.5 cos 3.77 t .
x = 0.5 cos 3.77 t .
5.1 m
Explanation:
Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by
(1)
where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground 
is given by
(2)
At the instant the two balls collide, they will have the same displacement, therefore
or
Solving for t, we get
We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):
Answer:
The lightbulb will NOT light.
Explanation:
You put me in a difficult position. I can't help it, but the "sample answer" is by far the best way to explain this, briefly and correctly. There's no other choice but to copy it.
This is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.
<em>If</em> the lower switch were <u>opened</u>, THEN we would have a series circuit. Current would no longer have any other choice but to flow through the bulb, and the bulb would light.
The red at the bottom is just the holder for the beaker. It’s not apart of the density.
Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
= magnitude of current
magnitude of current is given as
= 1.87 A
