Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 = 
ω2 = 216.47
<span>translational kinetic energy is larger than its rotational kinetic energy</span>
Answer:
0° C
Explanation:
Given that
Mass of ice, m = 50g
Mass of water, m(w) = 50g
Temperature of ice, T(i) = 0° C
Temperature of water, T(w) = 80° C
Also, it is known that
Specific heat of water, c = 1 cal/g/°C
Latent heat of ice, L(w) = 89 cal/g
Let us assume T to be the final temperature of mixture.
This makes the energy balance equation:
Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C
m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have
50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)
4000 + 50T = 4000 - 50T
0 = 100 T
T = 0° C
Thus, the final temperature is 0° C
Answer:
140m east
Explanation:
If East is positive then lets rephrase the problem into integers
A truck moves +70 m, then moves -120m, and finally moves +90m.
So totally Displacement = +70-120+90= +140m
Since east is positive, the trucks resultant displacement is 140 m east of origin
Answer:
It is 52° below the celestial equator.
Explanation:
The declination is the angle in degrees measured north (+) or south (-) of the an imaginary line called the celestial equator.
The celestial equator is a projection of the earth's equator on the celestial sphere. imaginary
The star named Canopus has a declination of approximately –52°.
Since the angle is negative, this shows that it is south or below the celestial equator and at 52° south of the celestial equator.
Thus, the star named Caponus is 52° below the celestial equator.
