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3 years ago

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Physical Education - Grade 9

1 answer:

Degger [83]3 years ago

5 0

you need to work harder to get your heart rate higher because you have become more physically fit in pace or you would be in the right heartrate

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Taking a timeout is effective as an anger management tool.

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Which of the following BEST describes a body in equilibrium?

sukhopar [10]

$\textsf {\large{\underline { \underline {SOLUTION:-}}}}$

⇰Let O be the point of observation on the ground OX.

⇰Let A and B be the two positions of the jet.

$\bf➠Then, \angle XOA=60⁰ \: and \: \angle \: XOB=30⁰ \\ \\$

$\bf ➠Draw \: AL \perp \: OX \: and \: BM \perp \: OX \\ \\$

$\bf \:⇰ Let \: AL=BM=h \: meters. \\ \\$

$\bf \: ⇰ \: Speed \: of \: Jet = 720 \: km/hr\\$

$\bf = \huge( \small \: 720 \times \frac{5}{18} \huge) \small \: m / s\\$

$\bf = 200 \: m/s$

$\bf \:➥ Time \: taken \: to \: cover \: the \: distance \: AB=15 \: sec \\ \\$

$\bf➥ Distance \: covered =(speed×time) \\ \\$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = (200 \times 15)m = 3000m \\ \\$

$\bf \therefore LM=AB=3000m \\ \\$

${ \bf \: ⇰ \: Let \: OL} = x \: \bf m \\ \\$

$\bf➥From \: right \: \triangle OLA, \: we \: have: \\$

$\bf \frac{OL}{AL} =cot \: 60⁰= \frac{1}{ \sqrt{3} } \implies \frac{x}{h} = \frac{1}{ \sqrt{3} } \\ \\$

$\bf \implies \: x = \frac{h}{ \sqrt{3} } \: \: \: \: \: \: \: ....(1) \\ \\$

$\bf \:➥ From \: right \: \triangle \: OMB,we \: have: \\$

$\bf\frac{OM}{BM} =cot30⁰= \sqrt{3} \\ \\$

$\bf \implies \: \frac{x + 3000}{h} = \sqrt{3} \\ \\$

$\bf[ \boxed{➯OM=OL+LM=OL+AB=(x+3000)m \: and \: BM=h \: m]}$

$\bf \implies \: x + 3000 = \sqrt{3} h \\$

$\bf \implies \: x = ( \sqrt{3} h - 3000) \: \: \: \: \: \: ....(2) \\$

⇰ Equating the value of x from (1) and (2),we get;

$\bf \: \sqrt{3} h - 3000 = \frac{h}{ \sqrt{3} } \\ \\$

$\implies \bf \: 3h - 3000 \sqrt{3} = h \\$

$\bf \implies \: 2h = (3000 \times \sqrt{3} ) = (3000 \times 1.762) \\$

$\bf \implies \: h = (3000 \times 0.866) = 2598 \\ \\$

$\bf \:➥ Hence \: ,the \: required \: height \: is \: \boxed{ \boxed{ \bf2598 \: m.}} \\ \\$

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I believe it is walking with a limp on the right side

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3.<br> Heat transfers without medium due to the process of<br> answer convection

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Answer:

<h2>Radiation</h2>

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