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amm1812
3 years ago
6

R o b l o x is down today

Physics
1 answer:
madam [21]3 years ago
6 0
nobody cares play a better game
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Electrical power is transmitted from power plants to consumers–sometimes over very long distances– through conducting power line
sergeinik [125]
There is more wire to travel through,farther distance, and a higher possibility of other disruptions.  Please Mark Brainliest!!!
7 0
3 years ago
It says find the slope for each line I'm stuck on number one can you help me
Allushta [10]
\text{slope}=\frac{y_2-y_1}{x_2-x_1}

(-2, -1)(3, 1)

Therefore,

\text{slope}=\frac{1-(-1)}{3-(-2)}=\frac{1+1}{3+2}=\frac{2}{5}

7 0
1 year ago
Which phenomena support only the wave theory of light? select 2 options. reflection refraction diffraction interference photoele
attashe74 [19]

Interference and diffraction are the phenomena that support only the wave theory of light. Options 2 and 3 are correct.

<h3 /><h3>What is the interference of waves?</h3>

The result of two or more wave trains flowing in opposite directions on a crossing or coinciding pathways. This phenomenon is known as the interference of waves.

The phenomenon of interference occurs when two wave pulses are traveling along a string toward each other.

The light wave hypothesis states that light behaves like a wave. Since light is an electromagnetic wave, it may be transmitted without a physical medium.

Light has magnetic and electric fields, much like electromagnetic waves do.

Transverse waves, such as those seen in light waves, oscillate in the same direction as the wave's path. A wave of light may experience interference as well as diffraction as a result of these properties.

All of the remaining options are the light phenomenon.

Hence, options 2 and 3 are correct.

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3 0
2 years ago
The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve
Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

\varepsilon _t=\dfrac{D-d}{d}

\varepsilon _t=\dfrac{32-30}{30}

\varepsilon _t=0.0667

We know that

n=-\dfrac{\epsilon _t}{\varepsilon _{long}}

\varepsilon _{long}=-\dfrac{\epsilon _t}{n}

\varepsilon _{long}=-\dfrac{0.0667}{0.45}

\varepsilon _{long}=-0.148

So the axial pressure

P=E\times \varepsilon _{long}

P=5\times 0.148

P=740 KPa

The movement in the sleeve

\Delta =\varepsilon _{long}\times L

\Delta =0.148\times 50

Δ=7.4 mm

6 0
3 years ago
A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com
bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature =20^{\circ}C

Final Temperature =80^{\circ}C

mass flow rate of cold fluid \dot{m_c}=1.2 kg/s

Initial Geothermal water temperature T_h_i=160^{\circ}C

Let final Temperature be T

mass flow rate of geothermal water \dot{m_h}=2 kg/s

diameter of inner wall d_i=1.5 cm

U_{overall}=640 W/m^2K

specific heat of water c=4.18 kJ/kg-K

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)

2\times (160-T)=1.2\times (80-20)

160-T=36

T=124^{\circ}C

As heat exchanger is counter flow therefore

\Delta T_1=160-80=80^{\circ}C

\Delta T_2=124-20=104^{\circ}C

LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}

LMTD=\frac{80-104}{\ln \frac{80}{104}}

LMTD=91.49^{\circ}C

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

\dot{m_c}c(80-20)=U\cdot A\cdot (LMTD)

A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2

A=\pi DL=5.144

L=\frac{5.144}{\pi \times 0.015}

L=109.16 m

6 0
3 years ago
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