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3 years ago

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Jack and Jill ran up the hill at 2.8 m/s . The horizontal component of Jill's velocity vector was 2.1 m/s . What was the angle o

f the hill?What was the vertical component of Jill's velocity?

1 answer:

klasskru [66]3 years ago

7 0

Answer: $\theta =41.409 ^{\circ}$

Explanation:

Given

Jack and Jill ran up the hill at 2.8 m/s

Horizontal component of Jill's velocity vector was 2.1 m/s

Let $\theta$ is the angle made by Jill's velocity with it's horizontal component

Therefore

$2.8cos\theta =2.1$

$cos\theta =\frac{2.1}{2.8}$

$cos\theta =0.75$

$\theta =41.409 ^{\circ}$

Vertical velocity is given by

$V_y=2.8sin41.11=1.85 m/s$

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Answer:

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b) $\frac{F_2}{L}=1.95*10^-^5N$

Explanation:

From the question we are told that:

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a)

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b)

Generally the equation for Force on $I_2$ due to $I_1$ is mathematically given by

$F_2=I_2B_1L$

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B_1=Magnetic field current by $I_2$

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$\frac{F_2}{L} =I_2(\frac{\mu *I_1*I_2}{2\pi d})$

$\frac{F_2}{L}=1.95*10^-^5N$

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