A metal of length 15.01m is heated until it temperature rises 60°(if it new length is 15.05cm, calculate it linear expansitivity
Answers
Answer:
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:Given
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01m
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘C
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequired
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivity
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:\alpha = \frac{L_2 - L_1}{\triangle T * L1}α=△T∗L1L2−L1
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:\alpha = \frac{L_2 - L_1}{\triangle T * L1}α=△T∗L1L2−L1\alpha = \frac{15.05m - 15.01m}{60^\circ C* 15.01m}α=60∘C∗15.01m15.05m−15.01m
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:\alpha = \frac{L_2 - L_1}{\triangle T * L1}α=△T∗L1L2−L1\alpha = \frac{15.05m - 15.01m}{60^\circ C* 15.01m}α=60∘C∗15.01m15.05m−15.01m\alpha = \frac{0.04m}{60^\circ C * 15.01m}α=60∘C∗15.01m0.04m
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:\alpha = \frac{L_2 - L_1}{\triangle T * L1}α=△T∗L1L2−L1\alpha = \frac{15.05m - 15.01m}{60^\circ C* 15.01m}α=60∘C∗15.01m15.05m−15.01m\alpha = \frac{0.04m}{60^\circ C * 15.01m}α=60∘C∗15.01m0.04m\alpha = \frac{0.04}{60^\circ C* 15.01}α=60∘C∗15.010.04
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:\alpha = \frac{L_2 - L_1}{\triangle T * L1}α=△T∗L1L2−L1\alpha = \frac{15.05m - 15.01m}{60^\circ C* 15.01m}α=60∘C∗15.01m15.05m−15.01m\alpha = \frac{0.04m}{60^\circ C * 15.01m}α=60∘C∗15.01m0.04m\alpha = \frac{0.04}{60^\circ C* 15.01}α=60∘C∗15.010.04\alpha = \frac{0.04}{900.6^\circ C}α=900.6∘C0.04
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:\alpha = \frac{L_2 - L_1}{\triangle T * L1}α=△T∗L1L2−L1\alpha = \frac{15.05m - 15.01m}{60^\circ C* 15.01m}α=60∘C∗15.01m15.05m−15.01m\alpha = \frac{0.04m}{60^\circ C * 15.01m}α=60∘C∗15.01m0.04m\alpha = \frac{0.04}{60^\circ C* 15.01}α=60∘C∗15.010.04\alpha = \frac{0.04}{900.6^\circ C}α=900.6∘C0.04\alpha = 0.00004441483^\circ C^{-1}α=0.00004441483∘C−1
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:\alpha = \frac{L_2 - L_1}{\triangle T * L1}α=△T∗L1L2−L1\alpha = \frac{15.05m - 15.01m}{60^\circ C* 15.01m}α=60∘C∗15.01m15.05m−15.01m\alpha = \frac{0.04m}{60^\circ C * 15.01m}α=60∘C∗15.01m0.04m\alpha = \frac{0.04}{60^\circ C* 15.01}α=60∘C∗15.010.04\alpha = \frac{0.04}{900.6^\circ C}α=900.6∘C0.04\alpha = 0.00004441483^\circ C^{-1}α=0.00004441483∘C−1\alpha = 4.441483 * 10^{-5} ^\circ C^{-1}
Answer:\alpha = 4.44 * 10^{-5} ^\circ C^{-1}Explanation:GivenL_1 = 15.01mL1=15.01mL_2 = 15.05mL2=15.05m\triangle T = 60^\circ C△T=60∘CRequiredDetermine its linear expansivityThis is calculated as:\alpha = \frac{L_2 - L_1}{\triangle T * L1}α=△T∗L1L2−L1\alpha = \frac{15.05m - 15.01m}{60^\circ C* 15.01m}α=60∘C∗15.01m15.05m−15.01m\alpha = \frac{0.04m}{60^\circ C * 15.01m}α=60∘C∗15.01m0.04m\alpha = \frac{0.04}{60^\circ C* 15.01}α=60∘C∗15.010.04\alpha = \frac{0.04}{900.6^\circ C}α=900.6∘C0.04\alpha = 0.00004441483^\circ C^{-1}α=0.00004441483∘C−1\alpha = 4.441483 * 10^{-5} ^\circ C^{-1}\alpha = 4.44 * 10^{-5} ^\circ C^{-1}