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3 years ago

12

What is the primary outgoing radiation put off by the sun?

1 answer:

ira [324]3 years ago

7 0

I really don’t know but I think it’s D

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During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial veloc

ale4655 [162]

Answer:

145.8m

Explanation:

The toss distance is given by:

$x=v_0^2*\frac{sin(2\phi)}{g} ,(g=9.8m/s^2)$

7 0

3 years ago

Why would knowing the characteristics of circuits be important in designing electrical circuits?

blondinia [14]

The characteristic of a circuit actually indicate how the circuit functions. When one designs a circuit they have a specific function in mind and must know how to combine components in order to fulfill this functions.

8 0

3 years ago

Read 2 more answers

The current and the potential difference in an inductor are in phase. B. The current lags the potential difference by π/2 in an

slava [35]

Answer:

The current lags the potential difference by π/2 in an inductor

Explanation:

The potential difference leads to the current by $\frac{\pi}{2}$ . Alternate signals such as current and voltage -in this case- are periodic, this means that this signals are repeated at fixed spaces of time. Thus, In an inductor the current lags the potential difference by $\frac{\pi}{2}$ .

6 0

3 years ago

A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of

levacccp [35]

$\underline {\huge \boxed{ \sf \color{skyblue}Answer : }}$

<u>Given :</u>

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{D} = 5mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{DSS} = 10mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS(off)} = -6V$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS} = {?}$

$\: \: \:$

<u>Let's Slove :</u><u> </u>

$\tt \large I_{D} = I_{(DSS)} (1 - \frac {V_{GS}}{V_{GS(off)}} )^{2}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times V_{GS(off)}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times { - 6}$

$\: \:$

$\underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}$

3 0

1 year ago

51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

5 0

3 years ago