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Katena32 [7]
3 years ago
10

Plz answer i need it. Physics wave.

Physics
1 answer:
WITCHER [35]3 years ago
4 0

Answer:

part \: \to a  \:  \\ \boxed{\gamma  = 1.17 { \times 10}^{ - 6}  \: m} \\  \\ part  \to\: b \\ \boxed{the \: ans = 355,102,030.67 \: or \: 3.55 { \times 10}^{8}  \: times}

Explanation:

\boxed{part \: a.} \\ let \: the \: radio \: wave \: length \: be \to  \gamma _{r} \:  \\ given \to \: v  = f \gamma _{r}  \:  \\ the \: wave \: length \:  \boxed{\gamma _{r}  } \: is \to \:  \frac{v}{f}  \\  \gamma _{r}   =  \frac{v}{f}  =  \frac{350}{3 { \times 10}^{8} }  = 1.1666667 { \times 10}^{ - 6} \\   \boxed{\gamma _{r}   = 1.17 { \times 10}^{ - 6}  \: m} \\  \\   \boxed{part \: b.}\\ let \: the \: water \: wave \: length \: be \to  \gamma _{w} \:   \\ to \: answer \: the \: second \: question : \\  first \: we \: find \: the\: frequency \: of \: the \\ \: water \: wave \to \\  \:  if \:  \to \: v = f\gamma _{w} \\ f =  \frac{v}{ \gamma _{w}}  \\ but \:\gamma _{w} \: is \: 1\% \: of \:  \gamma _{r} \\ \gamma _{w} =  \frac{1}{100}  \times 1.1666667 { \times 10}^{ - 6} \\ \gamma _{w} = \underline{  1.1666667 { \times 10}^{ - 8}}m \\ hence \to \\ f_{w} =  \frac{v}{ \gamma _{r}}  =  \frac{1450}{1.1666667 { \times 10}^{ - 8}} \\  = \boxed{part \: a.} \\ let \: the \: radio \: wave \: length \: be \to  \gamma _{r} \:  \\ given \to \: v  = f \gamma _{r}  \:  \\ the \: wave \: length \:  \boxed{\gamma _{r}  } \: is \to \:  \frac{v}{f}  \\  \gamma _{r}   =  \frac{v}{f}  =  \frac{350}{3 { \times 10}^{8} }  = 1.1666667 { \times 10}^{ - 6} \\   \boxed{\gamma _{r}   = 1.17 { \times 10}^{ - 6}  \: m} \\  \\   \boxed{part \: b.}\\ let \: the \: water \: wave \: length \: be \to  \gamma _{w} \:   \\ to \: answer \: the \: second \: question : \\  first \: we \: find \: the\: frequency \: of \: the \\ \: water \: wave \to \\  \:  if \:  \to \: v = f\gamma _{w} \\ f =  \frac{v}{ \gamma _{w}}  \\ but \:\gamma _{w} \: is \: 1\% \: of \:  \gamma _{r} \\ \gamma _{w} =  \frac{1}{100}  \times 1.1666667 { \times 10}^{ - 6} \\ \gamma _{w} = \underline{  1.1666667 { \times 10}^{ - 8}}m \\ hence \to \\ f_{w} =  \frac{v}{ \gamma _{r}}  =  \frac{1450}{1.1666667 { \times 10}^{ - 8}} \\  \boxed{f_{w} = 124,285,710,735} \\now \: thier \: frequency \: ratio \: is \to \\  \frac{f_{w}}{f_{r}}  =  \frac{124,285,710,735}{350}  = 355,102,030.67 \\  \boxed{the \: ans = 355,102,030.67 \: or \: 3.55 { \times 10}^{8}  \: times}

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Answer:

Ek1 = 900000 [J]

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In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

E_{k1}=\frac{1}{2}*m*v1^{2}

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Length of largest known mosquito in inches = ?

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In a tape recorder, the tape is pulled past the read-and-write heads at a constant speed by the drive mechanism. Consider the re
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Answer:

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Let it be α . Let I be moment of inertia of reel .

Reel is in the form of disc

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3 years ago
Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L − 1 and R max ( V max ) = 222 μmol ⋅ L − 1 ⋅ s − 1 , d
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Answer:

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Explanation:

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The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1

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Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1

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