Suppose that at room temperature, a certain aluminum bar is 1. m long. the bar gets longer when its temperature is raised. the length l of the bar obeys the following relation: l=1.+2.4×10−5t, where t is the number of degrees celsius above room temperature. what is the change of the bar's length if the temperature is raised to 13.0 ∘c above room temperature?

ΔL =0. 000312  m

Explanation:

Given that

At  room temperature ( T = 25 ∘C)  ,L= 1 m

So the length at  13.0 ∘C above room temperature

L=1.000312 m

So the change in length

ΔL = 1.000312 - 1.0000 m

ΔL =0. 000312  m

answer: 0

explanation: im   14

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft