Need help figuring this out keep getting the wrong answer. Let's look at the surface tension of the liquid metalmercury. Mercury has an unusually high surfacetension, about 465 dyn/cm Calculate the excesspressure inside a drop of mercury 4.00 mm indiameter.SOLUTIONSET UP AND SOLVE We first convert the surface tension y to Sl units:y= 465 dyn/cm = 465 x 10 3 N/m.Now we have(2)(465 x10-3 N/m)0.00200 m27p- PatmR465 Pa= 0.00459 atm.REFLECT The excess pressure (the difference between the inside and outside pressures) is avery small fraction of the atmospheric pressure.Part A - Practice Problem:What would the diameter of the drop be if the inside pressure were 5 atm?Express your answer in millimeters to three significant figures.?x•10"d =mmSubmitPrevious Answers Reguest Answer

Given: γ=465×10-3N/m, P=5 atm=5×105Pa The pressure difference inside and outside of the mercury drop…

Let's look at the surface tension of the liquid metal mercury. Mercury has an unusually high surface tension, about 465 dyn/cm. Calculate the excess pressure inside a drop of mercury 4.00 mm in SOLUTION SET UP AND SOLVE We first convert the surface tension y to Sl units: diameter. y= 465 dyn/cm = 465 x 10 N/m. Now we have 27 (2)(465 x 10-3 N/m) P- Patm R 0.00200 m = 465 Pa 0.00459 atm. REFLECT The excess pressure (the difference between the inside and outside pressures) is a very small fraction of the atmospheric pressure. Part A - Practice Problem: What would the diameter of the drop be if the inside pressure were 5 atm? Express your answer in millimeters to three significant figures. ΑΣφ ? x•10" d = mm Submit Previous Answers Request Answer

Let's look at the surface tension of the liquid metal mercury. Mercury has an unusually high surface tension, about 465 dyn/cm. Calculate the excess pressure inside a drop of mercury 4.00 mm in SOLUTION SET UP AND SOLVE We first convert the surface tension y to Sl units: diameter. y= 465 dyn/cm = 465 x 10 N/m. Now we have 27 (2)(465 x 10-3 N/m) P- Patm R 0.00200 m = 465 Pa 0.00459 atm. REFLECT The excess pressure (the difference between the inside and outside pressures) is a very small fraction of the atmospheric pressure. Part A - Practice Problem: What would the diameter of the drop be if the inside pressure were 5 atm? Express your answer in millimeters to three significant figures. ΑΣφ ? x•10" d = mm Submit Previous Answers Request Answer