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3 years ago

Which type of circuit has two or more branches for the current to flow through?

1 answer:

3 0

Hello,

The answer is A. Parallel circuit.

Unlike a series circuit, these circuits allow currents to flow through multiple branches. An open circuit is an electrical current that is not valid, therefore there will be no current flow. A combination circuit, I believe, <span>is a </span>circuit<span> that is a blend of series paths and parallel paths.

Faith xoxo</span>

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A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t

Shalnov [3]

Answer:

a) yield strength

$\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa$

b) modulus of elasticity

strain calculation

$\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046$

strain for offset yield point

$\varepsilon_{new} = \varepsilon_0 -0.002$

=0.0046-0.002 = 0.0026

now, modulus of elasticity

$E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}$

= 184615.28 MPa = 184.615 GPa

c) tensile strength

$\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa$

d) percentage elongation

$\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%$

e) percentage of area reduction

$\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%$

7 0

3 years ago

A material through which electrons can move easily is a what

kirill [66]

Answer:

Electrical conductors

3 0

1 year ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

7 0

3 years ago

Why are special techniques and extinguishing agents are required to fight combustible metals fires?

d1i1m1o1n [39]

Cucucicicicivigovogovphphpbpb

7 0

3 years ago

Why does a dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s?

andrew11 [14]

A dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.

s = vt - 1 / 2 at²

s = Displacement

v = Final velocity

t = Time

a = Acceleration

s = 5 m

t = 1 s

a = 10 m / s²

5 = ( v * 1 ) - ( 1 / 2 * 10 * 1 * 1 )

5 = v - 5

v = 10 m / s

The equation used to solve the given problem is an equation of motion. In a free fall motion, usually air resistance is not considered for easier calculation. If air resistance is considered acceleration cannot be constant throughout the entire motion.

Therefore, a dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.

To know more about equation of motion

brainly.com/question/5955789

#SPJ1

4 0

1 year ago