The specific heat of water is 4,186 J/kg.'C. Approximately how much heat must
1 answer:
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- T1=40°C=313K
- T_2=-10°C=263K
Applying Charles law
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration,
Diameter of the wheel, d=21 cm
Radius of wheel, cm
Radius of wheel,
1m=100 cm
Magnitude of total linear acceleration, a=
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,
Radial acceleration,
We know that
Using the formula
Squaring on both sides
we get
Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s