How many atoms are in SO<br> 3

A very long insulating cylinder of charge of radius 2.60 cm carries a uniform linear density of 15.0 nC/m . Part A If you put on

fenix001 [56]

Given Information:

Radius = ra = 2.60 cm = 0.026 m

Density = J = 15.0 nC/m

change in potential difference = ΔV = 200 V

Required Information:

Distance = d = ?

Answer:

distance = 0.088 m

Explanation:

As we know

ΔV = Vb - Va = J/4πε₀*ln(rb/ra)

Where ra and rb is the point where potential difference is Va and Vb respectively

1/4πε₀ = 9x10⁹ N.m²/C²

We want to find the distance d = rb - ra

ΔV = J/4πε₀*ln(rb/ra)

200 = 9x10⁹*15x10⁻⁹*ln(rb/ra)

200/135 = ln(rb/ra)

1.48 = ln(rb/ra)

taking e on both sides yields

e^(1.48) = rb/ra

4.39 = rb/ra

rb = 4.39*0.026

rb = 0.114 m

Therefore, the required distance is

d = rb - ra

d = 0.114 - 0.026

d = 0.088 m

Therefore, the other probe must be placed 0.088 m from the surface so that the voltmeter reads 200 V

6 0

3 years ago

Steam at 400C has a specific volume of 0.02m3/kg. Determine the pressure of the steam based on a) the ideal gas equation b) the

nikklg [1K]

Answer:

by ideal gas pressure = 15529.475 kPa

by compressibility chart pressure = 12576 kPa

by steam tables Pressure = 12517 kPa

Explanation:

given data

temperature T = 400°C = 673 K

volume v = 0.02 m³/kg

to find out

pressure by ideal gas, compressibility chart and steam tables

solution

we know here by table

gas constant R is 0.4615 kJ/ kg-K

and critical temp Tc = 647.1 K

and critical pressure Pc = 22064 kPa

so by ideal gas pressure is

pressure = R×T / v

pressure = 0.4615 × 673 / 0.02

pressure = 15529.475 kPa

and

by compressibility chart

temperature reduce is = T/ Tc

temperature reduce Tr = 673 / 647.1

Tr = 1.040 K

so pseudo reduce volume is here

reduce volume Vr = v / ( RTc/Pc)

reduce volume Vr = $\frac{0.02}{\frac{461.5(647.1)}{22064*10^{3} } }$

0.02 / ( 461.5(647.1) / 22064×10³)

reduce volume = 1.48

and we know by compressibility chart

reduce pressure Pr is 0.57

so

pressure = Pr × Pc

pressure = 0.57 × 22064 × 10³

pressure = 12576 kPa

and

from steam table

pressure is 12.5 MPa at 673 K and 0.020030 m³/kg

pressure is 15 MPa at 673 K and 0.015671 m³/kg

so

pressure P is

$\frac{0.02 - 0.020030}{0.015671 - 0.020030}$ = $\frac{ P - 12.5}{15 - 12.5}$

so

Pressure = 12517 kPa

4 0

3 years ago

A train is traveling at a speed of 50km/hr. How.many hours will it take the train to travel 800 km?

bagirrra123 [75]

Answer:

It would take 16 hours

Explanation:

800 divided by 50 = 16

Brainly pls

8 0

3 years ago

Read 2 more answers

An object starts from rest and accelerates to a velocity of 10 m/s, east in 5 seconds. What is the acceleration of the object?

Shalnov [3]

Answer:

2m/s²

Explanation:

acceleration is the rate of change of velocity

for velocity to reach 10m/s in 5s,

acceleration= 10/5

=2

3 0

4 years ago

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea

KonstantinChe [14]

Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = time passed by the sample = 90.3 min

a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

$k=\frac{2.303}{90.3min}\log\frac{400}{100}$

$k=1.54\times 10^{-2}\text{ min}^{-1}$

Now we have to calculate the half-life of substance, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$1.54\times 10^{-2}\text{ min}^{-1}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=45min$

Therefore, the half-life of this substance will be, 45 minutes.

4 0

3 years ago

How many atoms are in SO 3