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alexandr402 [8]
3 years ago
15

What is the magnitude of the linear momentum of a 7.30 kg bowling ball going down the

Physics
1 answer:
padilas [110]3 years ago
3 0
Momentum = mass x velocity
m=7.3kg
v=20 m/s

momentum= 7.3(20) = 146 kg m/s
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sweet-ann [11.9K]

\blue{\bold{\underline{\underline{Answer:}}}}

  • \green{\tt{Work\:done=24\:J}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

\green{\underline{\bold{Given :}}} \\  \tt: \implies Constant \: force(F) = 8 \: N \\  \\ \tt: \implies Displacement(s) = 3 \: m \\  \\ \red{\underline{\bold{To \: Find : }}} \\  \tt:  \implies Work \: Done(W.D) = ?

• <u>According to given question</u> :

\green{ \star} \tt \:  \theta \:  = 0 \degree \:  \:  \:  \: (Angle \: between \: force \: and \: displacement) \\  \\  \bold{As \: we \: know \: that} \\  \tt:  \implies Work \: Done = FS \: cos  \: \theta \\  \\  \tt:  \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\  \\ \green{ \circ} \tt \: cos  \: 0 \degree = 1  \\  \\  \tt:  \implies Work \: Done =24 \times 1 \\  \\   \green{\tt:  \implies Work \: Done =24 \: J}

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Explanation:

hope this helps tho i don't quite know what you mean

7 0
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F=qE=(8.1 \cdot 10^{-9}C)(9.4 \cdot 10^7 N/C)=0.76 N
5 0
3 years ago
Read 2 more answers
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