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1 year ago

Which of the following can be contracted from contact with bloodborne pathogens?

Physics

1 answer:

sergij07 [2.7K]1 year ago

6 0

HIV can be contracted from contact with bloodborne pathogens.

Other bloodborne diseases are HBV, malaria, syphilis and brucellosis

<h3>What are bloodborne pathogens?</h3>

Bloodborne pathogens can be defined as those microorganisms or pathogenic organisms that cause disease and are present in human blood.

Blood borne pathogens can also be contacted through the following means

Se- xual contact
Needle contact

In conclusion; HIV can be contracted from contact with bloodborne pathogens.

Learn more about bloodborne pathogens:

brainly.com/question/13158004

#SPJ1

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Hip hop dance translate feelings and emotions of human conditions, situations, or fantasies into movement and dramatic expressio

attashe74 [19]

Answer:

false .....................

4 0

1 year ago

A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height

lions [1.4K]

Answer:

(a) $47.15ms^{-1}$

(b) $2470.13ms^{-2}$

Explanation:

From Newton's second law of motion

F=ma where m is mass, a is acceleration and F is net force

Also from kinetic equation of motion, velocity and displacement are related using equation

$v^{2}=u^{2}+2sa$

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is $usin80^{o}$

Also, the vertical component can be written as

$v_{y}^{2} }=u_{y}^{2} } -2gs$ The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity, $u_{y}$ is vertical component of initial velocity and $v_{y}$ is vertical component of final velocity

Since $v_{y}$ is 0

$u_{y}^{2} } =2gs$

s=110 and g is taken as 9.8

$u_{y}=\sqrt{2*9.8*110}=46.43275$

$u_{y}=46.43ms^{-1}$

Also, it's evident that the vertical component of initial velocity is $u_{y}=u_{i}sin \theta$ where $\theta$ is angle of projection and $u_{i}$ is resultant velocity

Making $u_{i}$ the subject we obtain $u_{i}=\frac {u_{y}}{sin \theta}$

Since $u_{y}$ and $\theta$ are known as $46.43ms^{-1}$ and $80^{o}$ respectively, then $u_{i}=\frac {46.43ms^{-1}}{sin 80^{o}}=47.15ms^{-1}$

Therefore, the velocity of potato is $47.15ms^{-1}$

(b)

Displacement depends on length of tube hence s=0.450m hence going back to kinetic equation $v^{2}=u^{2}+2sa$

The final velocity v is answer in part a which is $47.15ms^{-1}$ , initial velocity u is $0ms^{-1}$ hence the equation is re-written as

$v^{2}=2sa$ and making a the subject we obtain

$a=\frac {v^{2}}{2s}$

$a=\frac {47.15^{2}}{2*0.450}=2470.13ms^{-2}$

Therefore, average acceleration is $2470.13ms^{-2}$

(c)

From Newton's second law of motion, F=ma where m=0.500kg and a is $2470.13ms^{-2}$

Therefore, the average force of potato is

F=0.5*2470.13=1235.06N

F=1235.06N

The weight, W of potato is given by W=mg

Taking R as ratio of average force and weight of potato

$R=\frac {F}{W}=\frac {F}{mg}$ and since F=1235.06, m=0.500kg and g=9.8

$R=\frac {1235.06}{0.500*9.8}$ =252.05

Therefore, ratio of average force to weight is 252.05

8 0

3 years ago

With full explaniation

zzz [600]

A. The ball's (vertical) velocity $v$ at time $t$ is

$v(t) = 30\dfrac{\rm m}{\rm s} - gt$

so that after 4 seconds, the ball's speed is

$|v(4\,\mathrm s)| = \left|30\dfrac{\rm m}{\rm s} - \left(10\dfrac{\rm m}{\mathrm s^2}\right) (4\,\mathrm s)\right| = \boxed{10\frac{\rm m}{\rm s}}$

(The velocity is -10 m/s, so the ball is falling back down at this point.)

B. At maximum height, the ball has zero velocity, so it takes

$30\dfrac{\rm m}{\rm s} - gt = 0 \implies t = \dfrac{30\frac{\rm m}{\rm s}}g = \boxed{3\,\mathrm s}$

for the ball to reach this height.

C. The height of the ball $y$ at time $t$ is

$y(t) = \left(30\dfrac{\rm m}{\rm s}\right) t - \dfrac g2 t^2$

The maximum height is attained by the ball at 3 seconds after it's thrown, so

$y_{\rm max} = \left(30\dfrac{\rm m}{\rm s}\right) (3\,\mathrm s) - \dfrac{10\frac{\rm m}{\mathrm s^2}}2 (3\,\mathrm s)^2 = \boxed{45\,\mathrm m}$

D. The time it takes for the ball to reach its maximum height is half the time it spends in the air. So the total airtime is $\boxed{6\,\mathrm s}$ .

Put another way: When the ball returns to the height from which it was thrown, its final velocity has the same magnitude as its initial velocity but points in the opposite direction. This is to say, after the total time the ball is in the air, it's final velocity will be -30 m/s. Then the total airtime is

$30\dfrac{\rm m}{\rm s} - gt = -30\dfrac{\rm m}{\rm s} \implies t = \dfrac{60\frac{\rm m}{\rm s}}g = \boxed{6\,\mathrm s}$