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3 years ago

11

Septum: Left and right atrium: Pacemaker: Left and right ventricles: Valves:

1 answer:

hammer [34]3 years ago

4 0

Can you define the question further.

This may help with your answer:

Tricuspid valve. This valve is located between the right atrium and the right ventricle.
Pulmonary valve. The pulmonary valve is located between the right ventricle and the pulmonary artery.
Mitral valve. This valve is located between the left atrium and the left ventricle. It has only 2 leaflets.
Aortic valve. The aortic valve is located between the left ventricle and the aorta

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Someone help me please

ivanzaharov [21]

C. cooked noodles and water
because noodles are long and water has no shape or size.
if you have any problems with this answer,
comment and I will fix it.
Thank s!

6 0

3 years ago

What is the speed of a wave with a wavelength of 3 m and a frequency of .1Hz?

Yuri [45]

We know,
Speed = Frequency * Wavelength
Speed = 3 * 0.1 m/s [hertz = 1/sec.]
So, your final answer is 0.3 m/s

Hope this helps!!

6 0

3 years ago

Read 2 more answers

How much kinetic energy the truck still has

seropon [69]

Answer:

47.5kJ

Explanation:

Before climbing the cliff

$E_t = E_k$

$E_k=\frac{1}{2} mv^2\\E_k=\frac{1}{2} 500*15^2\\\\E_k = 56250J\\$

At 2.

$E_t = E_k+E_p$

8 0

3 years ago

It requires a force of 100 N to stretch a spring of negligible mass by a distance of 0.1 m. If the spring instead stretches a di

Mariana [72]

by the formula of spring force we know that

$F = kx$

here we know that

$F = 100 N$

$x = 0.1 m$

now we will have

$100 = k \times 0.1$

$k = 1000 N/m$

now by similar way if the stretch in spring is 0.25 m

force is given by

$F = kx$

$F = 1000 \times 0.25$

$F = 250 N$

so it will require F = 250 N force

5 0

3 years ago

Read 2 more answers

Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De

My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0

3 years ago