1. A aluminum rod of density p = 2700 kg/m³ has a length of 3.0 m and cross sectional area 9.0 cm². It is bolted at both ends between two immobile supports. Initially, there is no tension in the rod because the rod just fits between the supports. Through cooling, the rod loses 3500 J of heat. a) Find the temperature change from the heat loss. b) Find the force on the rod that develops from the heat loss. 65 × 10⁹ Pa, a AL X Useful constants CAL = 910 J/(kgK), YAIL = = 22 x 10-6K-1.
1. A aluminum rod of density p = 2700 kg/m³ has a length of 3.0 m and cross sectional area 9.0 cm². It is bolted at both ends between two immobile supports. Initially, there is no tension in the rod because the rod just fits between the supports. Through cooling, the rod loses 3500 J of heat. a) Find the temperature change from the heat loss. b) Find the force on the rod that develops from the heat loss. 65 × 10⁹ Pa, a AL X Useful constants CAL = 910 J/(kgK), YAIL = = 22 x 10-6K-1.
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Transcribed Image Text:1. A aluminum rod of density p= 2700 kg/m³ has a length of 3.0 m and cross sectional
area 9.0 cm². It is bolted at both ends between two immobile supports. Initially, there is
no tension in the rod because the rod just fits between the supports. Through cooling, the
rod loses 3500 J of heat.
a) Find the temperature change from the heat loss.
b) Find the force on the rod that develops from the heat loss.
910 J/(kgK), YAL = 65 × 10⁹ Pa, a Al
=
Useful constants CAL
=
22 x 10-6K-1.
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