A 25.0 g sample of mercury is cooled from 50 °C to 10 °C. If the specific heat of mercury is 0.14 J/g x °C, calculate the amount of heat energy lost in the process.

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Ответ:

cuhh

19.06.2020

Physics

$\alpha =20.564^{o}$

Explanation:

Given data

Radius R=6.80 m

Velocity v=18 km/h =5 m/s

First we need to set up for Force in vertical (y) and horizontal (x)

∑Fy=0=TCosα-W

∑Fx=ma=Fc=TSinα

Solve Vertical force equation for T,substituting mg for W

$TCos\alpha -W=0\\TCos\alpha=W\\TCos\alpha=mg\\T=(mg/Cos\alpha)$

Substitute expression for Fc and T into horizontal force and simplify it we get

$Fc=TSin\alpha\\ where\\Fc=(mv^{2}/R )\\So\\(mv^{2}/R )=(mg/Cos\alpha )Sin\alpha\\(mv^{2}/R )=mg*tan\alpha\\ tan\alpha =\frac{v^{2} }{Rg}\\ \alpha =tan^{-1}(\frac{(5m/s)^{2} }{(6.80m)(9,8m/s^{2} )} )\\ \alpha =20.564^{o}$

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