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3 years ago

5

Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 C

/m.

1 answer:

Len [333]3 years ago

6 0

Answer: 71.93 *10^3 N/C

Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:

∫E*dr=Q inside/εo Q inside is given by: λ*L then,

E*2*π*r*L=λ*L/εo

E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C

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The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t

EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

$Ay = By = \frac{w * 6}{2} = 3w$

$P_c_r = 3w * F.S = 3w * 2.0 = 6w$

$I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7$

To find the maximum intensity, w, let's take the Pcr formula, we have:

$P_c_r = \frac{\pi^2 E I}{(KL)^2}$

Let's take k = 1

$E = 200*10^9$

Substituting figures, we have:

$6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}$

Solving for w, we have:

$w = \frac{67266.84}{6}$

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

$\sigma _c_r = \frac{w}{A}$

$\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y$ . This means it is safe

The maximum intensity w = 11.211KN/m

3 0

3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of 0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = <em>0.27</em>

7 0

3 years ago

Water emerges straight down from a faucet with a 2.51-cm diameter at a speed of 3.04 m/s. (because of the construction of the fa

Mkey [24]

This is a question on conservation of energy. That is,
mgh + KE1 = KE2
mgh +1/2mv1^2 = 1/2mv2^2
gh + 1/2v1^2 = 1/2v2^2

Where, h = 0.2 m, v1 =3.04 m/s
Therefore,
v2 = Sqrt [2(gh+1/2v1^2)] = Sqrt [2(9.81*0.2 + 1/2*3.04^2)] = 7.26 m/s

Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v
Where,
V = Av = πD^2/4*3.04 = π*(2.51/100)^2*1/4*3.04 = 1.504*10^-3 m^3/s

At 0.2 m below,
V = 1.504*10^-3 m^3/s = A*7.26
A = (1.504*10^-3)/7.26 = 2.072*10^-4 m^2

But, A = πr^2
Then,
r = Sqrt (A/π) = Sqrt (2.072*10^-4/π) = 0.121*10^-3 m
Diameter = 2r = 0.0162 m = 1.62 cm

3 0

3 years ago

A thin partition divides a thermally insulated vessel into a lower compartment of volume V and an upper compartment of volume 2V

Anni [7]

Answer:

1.89mol

Explanation:

The entropy change during free expansion is express as

$S_{f}-S_{i}=nRln(\frac{V_{F}}{V_{I}})\\$

Where S is the entropy of the system,

n is the amount of mole

R is the gas constant = 8.314 and

V is the volume occupied at the initial and final stage

since the process is n adiabatic free expansion, the entropy of the system is constant. Hence we can re-write the equation as

$S=nRln(\frac{V_{F}}{V_{I}})\\$

where the $V_{i}=v\\$ and $V_{f}=2v+v=3v\\$

$S=17.28J/k\\$ and

$R=8.314\\$

Now if we substitute in values we arrive at

$17.28=(8.314)n*ln(\frac{3v}{v} )\\17.28=(8.314)n*ln(3 )\\17.28=(8.314)n*1.0986\\n=\frac{17.28}{8.314*1.0989}\\n=1.89 mole\\$

6 0

3 years ago

A tsunami is ________. A. a sea wave generated by an earthquake, landslide, or meteor impact that may destroy coastal cities tho

a_sh-v [17]

Answer:

A. A sea wave generated by a displacement of water

Explanation:

7 0

3 years ago