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DIA [1.3K]
3 years ago
15

Which of the following is an example of how HIV can be transmitted from one person to another?

Health
1 answer:
Ksenya-84 [330]3 years ago
3 0

HIV can be transmitted from one person to another person by contact between infected saliva and an open wound.  

Further explanation:

Transmission of HIV:

A person can get HIV by coming into direct contact with certain body fluids from a person who is suffering from the disease.  

These body fluids include

  1. Blood
  2. Rectal fluid  
  3. Vaginal fluid  
  4. Breast milk
  5. Saliva

When these fluids having HIV come in contact with the bloodstream of a person who is HIV negative through a mucous membrane ( which is present in the vagina, mouth, rectum, or on the tip of the penis ); open sore or cut, or by direct injection.  

Activities by which HIV spread from person to person:

HIV spread by specific activities. In the United States, HIV is commonly spread by following method:

Sex with a person having HIV. Sharing the injection drug equipments for example needles with a person having HIV.  

While the less common methods of spread of HIV are

From a mother to a child during birth, pregnancy or breast feeding. The use of HIV medicines has lowered the risk of transmission of HIV from mother to child in the United States.  

Answer details

Subject: Health

Level: High school

Keywords

  • Transmission of HIV  
  • Activities by which HIV spread from person to person

Learn more to evaluate

brainly.com/question/6393374

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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}  \\  \\

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

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

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