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4 years ago

Solve the following automotive-services problem.

Physics

1 answer:

anyanavicka [17]4 years ago

7 0

Answer:

voltage =6V (T)

Explanation:

I= E/R

This formular is derived from ohms law

where E stands for Electro motive force measured in volts and often used interchangeably with voltage and measured in volts

I stands for current and R represents resistance to current flow

E=I × R

E= 3O × 0.2

E= 6V

which makes 6 volts true

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Please write in complete sentences.

BabaBlast [244]

Answer:

The density of a material affects the speed that a wave will be transmitted through it. In general, the denser the transparent material, the more slowly light travels through it.

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3 years ago

Are the stack temperature and oxygen reasonable for these operating conditions? if not, what oxygen and stack temperature would

Anton [14]

Stack temperatures typically range from 350 to 450 degrees Fahrenheit. A 2.5% efficiency loss occurs for every 100 degrees over that temperature. The majority of buildings schedule annual boiler cleanings at regular intervals, but if you see those figures rise, it's time for a cleaning.

Excess air is required to completely burn the fuel since the air and fuel cannot combine exactly in a burner. Additionally, any leaks in the heater will draw air into the firebox that doesn't pass through the burners since the furnace or boiler firebox operates at a little negative gauge pressure. Fuels that are gaseous, like natural gas, burn more readily than fuels that are liquid or solid. Depending on the fuel type, different surplus air requirements will apply.

Learn more about temperature here-

brainly.com/question/15267055

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3 0

1 year ago

It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet

TEA [102]

Answer:

$v=346.05\ m.s^{-1}$

Explanation:

Given:

initial temperature of the lead bullet, $T_i=43^{\circ}C$

latent heat of fusion of lead, $L_f=2.32\times 10^4\ J.kg^{-1}$

melting point of lead, $T_m=327.3^{\circ}C$

We have:

specific heat capacity of lead, $c=129\ J.kg^{-1}.K^{-1}$

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

$\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)$

$\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f$

$\frac{1}{2} m.v^2=m(c.\Delta T+L_f)$

$\frac{v^2}{2} =129\times(327.3-43)+23200$

$v=346.05\ m.s^{-1}$

3 0

3 years ago

Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.

romanna [79]

Answer:

(a): $\rm -5.627\times 10^3\ N.$

(b): $\rm 7.626\times 10^2\ N.$

Explanation:

<u>Given:</u>

Charge on one sphere, $\rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.$

Charge on second sphere, $\rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.$
Separation between the spheres, $\rm r=3.59\ cm = 3.59\times 10^{-2}\ m.$

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}$

where,

k is called the Coulomb's constant, whose value is $\rm 9\times 10^9\ Nm^2/C^2.$

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.$

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, $\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.$

The new charges on both the spheres are equal and given by

$\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.$

The magnitude of the force that each sphere now experiences is given by

$\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.$

7 0

3 years ago

A timline is necessary :

kolezko [41]

Answer:

A. For every goal

Explanation:

A timeline is necessary for every goal set, be it academic, life, high school, personal, and the like. A timeline is an important part of goal setting. It helps determine what should be done by a certain time to ensure that you will reach your goal. Although some goals can take a long time to reach, a timeline will be your constant reminder of what needs to be done.

When setting a timeline, you also need to take in consideration what you want to achieve. You need to set a realistic timeline as many usually underestimate the obstacles they may face in pursuing their goals.

8 0

4 years ago