The uncertainty principle applies not only to momentum and position, but also to energy and time.
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The answer:
the full question is as follow:
<span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D?
As shown in the figure,
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
D = 0 - (A + B + C) ,
let W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics
Magnitude
A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
the direction of D can be found by using Dx and Dy value
we know that tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°
direction is 26.02°
the full question is as follow:
<span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D?
As shown in the figure,
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
D = 0 - (A + B + C) ,
let W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics
Magnitude
A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
the direction of D can be found by using Dx and Dy value
we know that tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°
direction is 26.02°
π=iMRT
Where, π is Osmotic pressure,
i=1 for non-electrolytes,
M is molar concentration of dissolved species (units of mol/L)
R is the ideal gas constant = 0.08206 L atm mol⁻¹K⁻¹,
T is the temperature in Kelvin(K),
Here, to calculate M convert into standard units mg tog, ml to L, c to Kelvin
M= ( *10⁻³ )/ 0.175 =(5.987 *10⁻⁵)mol / 0.175L = 34.21*10⁻⁵ mol/L
π=iMRT=(1)*(34.21*10⁻⁵)*(0.08206)*(298.15)=837×10⁻⁵= 8.37×10⁻³ atm
=6.36 torr
(1 atm=760 torr, 1 Kelvin =273.15 °C, 1L=1000ml, 1g=1000mg)