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2 years ago

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.

Physics

1 answer:

Airida [17]2 years ago

4 0

Answer:

Part A : E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B : V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

Part A

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

F = $\frac{1}{4\pi}$ ε₀ Q₁/R₁²

Then the electric field at that point is

E = F/1

E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

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With countercurrent flow, diffusion happened in all regions of the filter. Explain why

jeka94

Answer:

When the blood and the dialysate are flowing in the same direction, as the the dialysate and the blood move away from the region of higher concentration of the urea, to a region distant from the source, the concentration of urea in the blood stream and in the dialysis reach equilibrium and diffusion across the semipermeable membrane stops within the higher filter regions such as II, III, IV or V

However, for counter current flow, as the concentration of the urea in the blood stream becomes increasingly lesser the, it encounters increasingly unadulterated dialysate coming from the dialysate source, such that diffusion takes place in all regions of the filter

Explanation:

3 0

3 years ago

A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s

BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

$Sf^2=So^2+(2*a*x)$

$Sf=So+(a*t)$

We are going to put our data:

$(15m/s)^2=So^2+(2*a*60m)$

$15m/s=So+(a*6s)$

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

$\sqrt{(15m/s)^2-(2*a*60m)}=So$

$15m/s-(a*6s)=So$

$\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)$

$[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}$

$(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}$

$-120m*a=-180m*a+36s^{2}*a^{2}$

$0=120m*a-180m*a+36s^{2}*a^{2}$

$0=-60m*a+36s^{2}*a^{2}$

$0=(-60m+36s^{2}*a)*a$

$0=a1$

$\frac{60m}{36s^{2}} = a2$

$1.67m/s^{2}=a2$

If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

$Sf=So+(a*t)$

$15m/s=So+1.67m/s^2*6s$

$15m/s-(1.67m/s^2*6s)=So$

$4.98m/s=So$

For part c:

We are going to use:

$Sf^2=So^2+(2*a*x)$

$(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)$

$\frac{24.80m^2/s^2}{3.34m/s^2}=x$

$7.42m=x$

5 0

2 years ago

A tuning fork produces a sound with a frequency of 256 hz and a wavelength in air of 1.33 m. find the speed of sound in the vici

aalyn [17]

We can think this through intuitively. A frequency of 256 Hz means that the wave has 256 cycles each second. If the wavelength is 1.33 meters, then there are 256 of them each second. Therefore, we just need to multiply the wavelength by the frequency to find the speed of sound. (Note that the units Hz = 1 / s) v = (frequency) x (wavelength) v = (256 Hz) x (1.33 m) v = 340.5 m/s The speed of sound in the vicinity of the fork is 340.5 m/s

4 0

2 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

Juliette [100K]

Answer:

24.348mm

Explanation:

NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards

K = d / €^n

Note : d represents the greek alphabet epsilion.

K = 345 / 0.02⁰.²² = 816mPa

The true strain based upon the stress of 414mPa =

€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576

However the true relationship between true strain and length is given by

€ = ln(Li/Lo)

Making Li the subject of formula by rearranging,

Li = Lo.e^€

Li = 520e⁰.⁰⁴⁵⁷⁶

Li = 544.348mm

The amount of elongation can be calculated from

Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.

8 0

2 years ago

If you find an igneous rock which has 450 radioactive isotopes and 3,150 stable daughter isotopes, how many half-lifes of this i

slavikrds [6]

Answer:

$3t_{1/2}$

Explanation:

To find the half-lifes of the isotope we need to use the following equation:

$N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}}$ (1)

where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and $t_{1/2}$ : is the half-lifes.

By solving equation (1) for t we have:

$\frac{t}{t_{1/2}} = - \frac{Ln(Nt/N_{0})}{Ln(2)}$

Having that:

Nt = 450

N₀ = 3150 + 450 = 3600,

The half-lifes of the isotope is:

$t = - \frac{Ln(450/3600)}{Ln(2)} \cdot t_{1/2} = 3t_{1/2}$

Therefore, 3 half-lives of the isotope passed since the rock was formed.

I hope it helps you!

3 0

2 years ago