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3 months ago

Smoking increases ____ levels in the blood, thus increasing the possibility of unwanted clotting.

1 answer:

7 0

It may lead to higher levels of homocysteine, which can harm the inner linings of arteries. Such damage often results from unwanted clotting that can occur due to factors like smoking, which tends to raise unwanted blood clots in the body, ultimately causing these levels to rise and resulting in undesired consequences.

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A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller piston. If the smaller piston has an a

kicyunya [3294]

Answer:

The cross-sectional area of the larger piston is 392cm ^{2}[/tex]

Explanation:

To find the solution, we apply the following equation:

Pascal's principle: F=P*A Formula (1)

F=Force applied to the piston

P: Pressure

A= Area of the piston

Nomenclature:

Fp= Force on the primary piston= 500N

W= weight of the car =m*g=2000kg*9.8m/s2= 19600N

Fs= Force on the secondary piston= W = 19600N

$Ap= Primary piston area=10cm^{2} =10*10^{-4}m^{2}$

As= Area of the secondary piston=?

Pressure applied on one side is distributed to all liquid molecules since liquids are incompressible.

From equation (1)

P=F/A

Pp=Ps

$\frac{Fp}{Ap} = \frac{Fs}{As}$

$As= \frac{Fs*Ap}{Fp}$

$As=\frac{19600*10*10^{-4} }{500}$

$As=0.0392m^{2} =0.0392*10^{4}cm^{2}$

$As=392cm ^{2}$

5 0

3 months ago

A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer

Keith_Richards [3271]

Response:

Clarification:

Provided

weight of disk $m=12.5 kg$

diameter of disc $R=0.23 m$

weight of ring $m_r=7 kg$

Force $F=9.7 N$

$N=180 rpm$

$\omega =\frac{2\pi N}{60}$

$\omega =6\pi rad/s$

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

$=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2$

$=13.25\times 0.23^2=0.7009 kg-m^2$

At this point, Torque is $T=F\times R=I\cdot \alpha$

$9.7\times 0.23=0.7\times \alpha$

$\alpha =3.18 rad/s^2$

Utilizing $\omega _f=\omega +\alpha t$

$\omega _f=0$ in this scenario

$0=6\pi -3.18\times t$

$t=\frac{6\pi }{3.18}$

$t=5.92 s$

7 0

2 months ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

serg [3582]

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁

7 0

3 months ago