The transfer of energy through matter by the direct contact of particles is called conduction
Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
<u>α = -0.16 rad/s²</u>
<u>Negative sign shows deceleration</u>
<u></u>
(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
<u>t = 33.2 s</u>
Well, first of all, a car moving around a circular curve is not moving
with uniform velocity. The direction of motion is part of velocity, and
the direction is constantly changing on a curve.
The centripetal force that keeps an object moving in a circle is
Force = (mass of the object) · (speed)² / (radius of the circle)
F = m s² / r
We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.
F = m s² / r
Multiply each side by 'r': F· r = m · s²
Divide each side by 'F': r = m · s² / F
We know all the numbers on the right side,
so we can pluggum in:
r = m · s² / F
r = (1200 kg) · (20 m/s)² / (6000 N) .
I'm pretty sure you can finish it up from here.
Distance = Speed × time
Distance = 31m/s × 13 seconds = 403m
Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
