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Delicious77 [7]
3 years ago
15

Which action will reduce the risk risk of injury on a ATV

Physics
2 answers:
sasho [114]3 years ago
4 0
Ur not getting on the ATV. I hope this helps
monitta3 years ago
4 0

Answer:

Helmet use, safe riding behaviours, restricting ridership by young operators etc. can reduce risk of injury on a ATV

Explanation: ATV stands for All-terrain vehicles. These are widely used in Canada for transportation, recreation, and occupations such as farming etc. It can be dangerous especially when it is used by the young adolescents and children who lacks strength, knowledge, and motor skills to operate them in a safe manner.

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Matter that organisms require for their life processes are
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nutrients

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Nutrients are matter that organisms require for their life processes.

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a sample of iron has the dimensions of 2cm times 3cm times 2cm. if the mass of this rectangular-shaped object is 94g, what is th
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The answer is 7.83. The three is repeating.

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What happens when exhale
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The current required to stimulate the heart during ventricular fibrillation is about 110mA (1000mA = 1A). Assuming that a conduc
algol [13]

Answer:

Explanation:

Given

current required I=110 mA

Internal Resistance R=300 \Omega

According to ohm law

Current flows in a conductor is directly Proportional to the voltage applied.

V\propto I

V=IR

V=110\times 10^{-3}\times 300

V=33 V  

5 0
3 years ago
A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is
Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12  m

Let assume the tank have a thickness  = dh

The radius of the tank by using the concept of similar triangle is :

\dfrac{1}{r} = \dfrac{12}{h}

r = \dfrac{h}{12}

The area of the tank = \mathbf{\pi r^2}

The area of the tank = \mathbf{\pi( \dfrac{h}{12})^2}

The area of the tank = \mathbf{ \dfrac{\pi}{144}h^2}

The volume of the tank is  = area × thickness

= \mathbf{ \dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \rho_ g * volume

where;

\rho_g = density of water ; which is given as 10000 N/m³

So;

Weight of the element = \mathbf{ 10000 *\dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \mathbf{69.44 \ \pi  \ h^2 \  dh}

However; the work required to pump this water = weight × height  rise

where the height rise = 12 - h

the work required to pump this water  = \mathbf{69.44 \ \pi  \ h^2 \  dh}(12 - h)

the work required to pump this water  = \mathbf{69.44 \pi (12h^2-h^3)dh}

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = \mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}

= \mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}

=  \mathbf{ 69.44 \pi[ \frac{12h^3}{3}-  \frac{h^4}{4}]^{12}}_0} }

= \mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}

= \mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}

= \mathbf{69.44 \pi*12^4 *\frac{1}{12}}

= 376966.991 Joules

6 0
3 years ago
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