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anastassius [24]
3 years ago
5

A 300.0-kg speedboat is moving across a lake at 35.0 m/s. what is the momentum of the speedboat?

Physics
1 answer:
alexandr402 [8]3 years ago
3 0
The momentum (n) of object is the product its mass (m) and velocity (v).
                                      n = m x v
Substituting the known values, 
                                      n = (300 kg) x (35 m/s) 
Thus, the value of the momentum is 10,500 kg m/s.
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5 0
3 years ago
Bob is pushing a box across the floor at a constant speed of 1.5 m/s, applying a horizontal force whose magnitude is 60 n. alice
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7 0
3 years ago
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5 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
In tokyo ghoul book 2 who is the main character and who taught ken kaneki the kagune
Inessa05 [86]

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Explanation: if I'm wrong correct me and i'll do more research

3 0
3 years ago
Read 2 more answers
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