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Gre4nikov [31]
2 years ago
12

The viscosity of all liquids decreases as the temperature is a) Increased b) decreased c) maintained constant d) fluctuating e)

none of these
Engineering
1 answer:
mars1129 [50]2 years ago
5 0

Answer:

b) decreased

Explanation:

Viscosity is the measure of the resistance between the layers of the liquid. On increasing temperature, the kinetic energy of the molecules of the liquid increases which in turn increase in the random motion of the molecules. Thus,  the relative motion of  the molecules in the liquid become easier and hence, viscosity decreases.

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A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po
Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: 4.3\cdot 10^{-15} J

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

V(r)=\frac{kq}{r}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

q_1=+2.00\mu C=+2.00\cdot 10^{-6}C

and is located at the origin (x=0, y=0)

Charge 2 is

q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

r_{1A}=0.40 m

So the potential due to charge 2 is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V

The distance of point A from charge 2 is

r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m

So the potential due to charge 1 is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V

Therefore, the net potential at point A is

V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m

So the potential due to charge 1 at point B is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V

The distance of charge 2 from point B is

r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m

So the potential due to charge 2 at point B is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V

Therefore, the net potential at point B is

V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V

So the potential difference is

V_B-V_A=-29,500 V-(-2700 V)=-26,800 V

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

W=q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

In this problem, we have:

q=-1.6\cdot 10^{-19}C is the charge of the electron

\Delta V=-26,800 V is the potential difference

Therefore, the work required on the electron is

W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J

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Answer: Borderline personality Disorder.

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Its symptoms includes unstable emotions, sense of insecurity, worthlessness, and impulsivity.

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Imagine you compare the effectiveness of four different types of stimulant to keep you awake while revising statistics using a o
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Answer:

The correct answer is At least two of the stimulants will have different effects on the mean time spent awake.

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