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3 years ago

Shotguns are examples of

Physics

1 answer:

tankabanditka [31]3 years ago

6 0

The answer would be {B}

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PLZ help asap :-/<br>............................

Misha Larkins [42]

Explanation:

$\underline{\boxed{\large{\bf{Option \; A!! }}}}$

Here,

$\rm { R_1}$ = 2Ω
$\rm { R_2}$ = 2Ω
$\rm { R_3}$ = 2Ω
$\rm { R_4}$ = 2Ω

We have to find the equivalent resistance of the circuit.

Here, $\rm { R_1}$ and $\rm { R_2}$ are connected in series, so their combined resistance will be given by,

$\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\$

$\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\$

$\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\$

Now, the combined resistance of $\rm { R_1}$ and $\rm { R_2}$ is connected in parallel combination with $\rm { R_3}$ , so their combined resistance will be given by,

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\$

Reciprocating both sides,

$\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\$

Now, the combined resistance of $\rm { R_1}$ , $\rm { R_2}$ and $\rm { R_3}$ is connected in series combination with $\rm { R_4}$ . So, equivalent resistance will be given by,

$\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\$

$\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\$

Henceforth, Option A is correct.

$\underline{\boxed{\large{\bf{Option \; B!! }}}}$

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

$\longrightarrow$ V = IR

$\longrightarrow$ 3 = I × 3.33

$\longrightarrow$ 3 ÷ 3.33 = I

$\longrightarrow$ 0.90 Ampere = I

Henceforth, Option B is correct.

$\tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\$

4 0

3 years ago

You were driving your car to UTD at a speed of 35 miles per hour. You stopped at the FloydCampbell intersection with the signal

ser-zykov [4K]

Answer:

green light have high energy

Explanation:

We have given the wavelength of the red light $\lambda =6.45\times 10^{-5}cm=6.45\times 10^{-7}m$

Speed of the light $c=3\times 106{8}m/sec$

The energy of the signal is given by $E=h\nu =h\frac{c}{\lambda }=\frac{6.67\times 10^{-34}\times 3\times 10^{8}}{6.45\times 10^{-7}}=3.1023\times 10^{-15}j$