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irakobra [83]
3 years ago
15

Which is an acceptable method to use when trying to stop severe bleeding? A. Apply pressure to a main artery leading to the woun

d. B. Place a clean cloth over the wound and press firmly with your hand. C. To slow blood flow, elevate the wound above the level of the heart. D. All of the above.
Health
2 answers:
Zolol [24]3 years ago
7 0
I'm fairly sure the answer is A
____ [38]3 years ago
3 0
It is all of the above a and b are the same thing basically and c just makes sense to do.
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Which of the following BEST describes a body in equilibrium?
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\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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