Sample Response: "No, steel and carbon would not form metallic bonds because metallic bonds only form between metals. Iron is a metal, but carbon is not."
To solve the problem it is necessary to take into account the concepts related to simple pendulum, i.e., a point mass that is suspended from a weightless string. Such a pendulum moves in a harmonic motion -the oscillations repeat regularly, and kineticenergy is transformed into potntial energy and vice versa.
In the given problem half of the period is equivalent to 1 second so the pendulum period is,

From the equations describing the period of a simple pendulum you have to

Where
g= gravity
L = Length
T = Period
Re-arrange to find L we have

Replacing the values,


In the case of the reduction of gravity because the pendulum is in another celestial body, as the moon for example would happen that,




In this way preserving the same length of the rope but decreasing the gravity the Period would increase considerably.
Answer:
Explanation:
Mass of car (M)=1200kg
Initial velocity (u)=20m/s
Stop after time (t)=3sec.
Come to stop implies that the final velocity is zero, v=0m/s
Using newton second law of motion
F=m(v-u)/t
Ft=m(v-u)
Since impulse is Ft
I=Ft
Then, I=Ft=m(v-u)
I=m(v-u)
I=1200(0-20)
I=1200×-20
I=-24,000Ns
The impulse delivered to the car by static friction is -24,000Ns
<span>Answer:
KE = (11/2)mω²r²,
particle B must have mass of 2m, while A has mass m.
Then the moment of inertia of the system is
I = Σ md² = m*(3r)² + 2m*r² = 11mr²
and then
KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2
So I'll proceed under that assumption.
For particle A, translational KEa = ½mv²
but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r²
For particld B, translational KEb = ½(2m)v²
but v = ω*r, so KEb = ½(2m)ω²r²
so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2
which is equal to our rotational KE.</span>