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3 years ago

5

A 11.8-m-long steel [E = 206 GPa] pipe column has an outside diameter of 202 mm and a wall thickness of 5 mm. The column is supp

orted only at its ends and it may buckle in any direction. Calculate the critical load Pcr for the following end conditions:

1 answer:

Nostrana [21]3 years ago

7 0

Answer:

A

Explanation:

i took the test pimp

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There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

3 years ago

If you wanted to

VLD [36.1K]

Answer:

Option C

Explanation:

According to the formula

$\\ \boxed{\sf R=\rho\dfrac{\ell}{A}}$

So

If we use wide wire we increase the area of cross section so resistance decreases

5 0

2 years ago

The early workers in spectroscopy (Fraunhofer with the solar spectrum, Bunsen and Kirchhoff with laboratory spectra) discovered

Anestetic [448]

The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.

<h3><u>Explanation: </u></h3>

A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.

Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen. "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.

The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.

8 0

3 years ago

Is the star with the longest total lifetime Also the farthest from earth? Explain.

Stels [109]

E) No. Ollie will shine for 30 Billion years but is only 10,000 LY from Earth.
F) No. Cosmo will shine for 3 Million years but is 10 Billion LY from Earth.
G) Yes. Ollie is only 10.000 LY away but will shine for 30 Billion years.
Ga) No. Stars such as Cosmo shine for 3 Million years.
Gb) If Cosmo was also 3 Million LY away we would see it now.

6 0

3 years ago

In a large restaurant an average 60% customers ask for water with their meal. A random sample of 10 customers is selected. Find

Gekata [30.6K]

Answer:

a) $P(6)=0.25$

b) $p(x$

c) $p(x\geq3)=0.9878$

d) $\sigma=\sqrt{2.4}=1.5492$

Explanation:

From the question we are told that:

Population percentage $p_\%=\60%$

Sample size $n=10$

Let x =customers ask for water

Let y =customers dose not ask for water with their meal

Generally the equation for y is mathematically given by

$y=1-p_\%\\y=1-0.60\\y=0.40$

Generally the equation for pmf p(x) is mathematically given by

$P(x)=10C_x (0..6)^x(0.4)^{10-x}$

a)

Generally the probability that exactly 6 ask for water is mathematically given by

$P(x)6=10C_6 (0..6)^6(0.4)^{10-6}$

$P(6)=0.25$

b)

Generally the probability that less than 9 ask for water with meal is mathematically given by

$p(xg)$

$p(x$

$p(x$

$p(x$

c)

Generally the probability that at least 3 ask for water with meal is mathematically given by

$p(x\geq3)=1-p(x$

$p(x\geq3)=1-[p(0)+p(1)+p(2)]$

$p(x\geq3)=1-[0.00001+0.0015+0.0106]$

$p(x\geq3)=1-[0.0122]$

$p(x\geq3)=0.9878$

d)

Generally the mean and standard deviation of sample size is mathematically given by

Mean

$\=x=np=10(0.6)=6$

Standard deviation

$v(x)=npq=10(0.6)(0.4)=2.4$

$\sigma=\sqrt{2.4}=1.5492$

4 0

3 years ago