7.

What is the total resistance of two 26 ohm resistors in series?

amid [387]

Answer:

I finish that module I'm sure it's D

4 0

3 years ago

Read 2 more answers

On a straight horizontal track along which blocks can slide with negligible friction, block 1 slides toward block 2, which is in

vivado [14]

If the mass and post-collision speed of block 1 is known, the momentum of the block 2 can be determined.

<h3>What is momentum?</h3>

The momentum of an object in motion is the product of mass and speed of the object.

P = mv

<h3>Conservation of linear momentum</h3>

The principle of conservation of linear momentum states that, the sum of the initial momentum must be equal to the sum of the final momentum.

$m_1v_1 = m_2 v_2$

Thus, if the mass and post-collision speed of block 1 is known, the momentum of the block 2 can be determined.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

7 0

2 years ago

How much electromagnetic energy is contained in each cubic meter near the Earth's surface if the intensity of sunlight under cle

WITCHER [35]

Answer:

344.8 x10^-8J/m³

Explanation:

Using=> energy intensity/ speed oflight

= 1000/2.9x10^8

= 344.8 x10^-8J/m³

3 0

3 years ago

How do the nonliving parts of earth's systems provide the basic materials to support life?

MatroZZZ [7]

The abiotic community of the ecosystem or the nonliving objects in the earth's biosphere influences the survival, growth and development of the organism. They provide basic mmaterials to support life by the biogeochemical processes which is composed of many biological and environmental cycles which profoundly has factored the intiation and continuation of life on earth. They also provide the basic materials to support life by the undergoing process which is also initiated by the living creatures or the biotic society of the ecosystem.

3 0

3 years ago

Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating wa

iren2701 [21]

Answer:

A. 6.36 lbm/s

b. $T_2=341\textdegree F$

Explanation:

a. Given the following information;

#Compressor inlet:

Air pressure, $P_1=14.7psia,T_1=60\textdegree F$

#Compressor outlet:

$Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min$

#Cooling rate,

$q_{out}=10Btu/lbm, \dot W=700hp$

# From table A-1E

Gas constant of air $R=0.3704\ psia.ft^3/lbm.R$

Specific enthalpy at $P_1=520R-h_1=124.27Btu/lbm$

Using the mass balance:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s$

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\\dot W_{in}+ \dot mh_1=\dot Q_{out}+\dot m_2\\\\h_2=\frac{\dot W_{in}-\dot Q_{out}}{\dot m}+h_1\\\\h_2=\frac{700\times 0.7068-10\times 6.36}{6.36}+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F$

Hence, the temperature at the compressor exit $T_2=341\textdegree F$

5 0

4 years ago