Ask question

Ask question

All categories

3 years ago

7

What part of the diagram is labeled by the number 3?

1 answer:

lutik1710 [3]3 years ago

5 0

Answer:

The part of the diagram thta is labeled number 3 is D. bronchi.

Explanation:

Bronchi are the airways that lead from the trachea to the lungs and they branch off into progressively smaller parts.

You might be interested in

Two people are pushing on a desk to the right with a force 15n each what is the net force on the desk

dybincka [34]

30N to the right direction

3 0

2 years ago

You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul

photoshop1234 [79]

1) First of all, let's find the resistance of the wire by using Ohm's law:
$V=IR$
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
$R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega$

2) Now that we have the value of the resistance, we can find the resistivity of the wire $\rho$ by using the following relationship:
$\rho = \frac{RA}{L}$
Where A is the cross-sectional area of the wire and L its length.
We already have its length $L=2.90 m$ , while we need to calculate the area A starting from the radius:
$A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2$

And now we can find the resistivity:
$\rho = \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}= 1.48 \cdot 10^{-7}\Omega \cdot m$

7 0

3 years ago

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

2 years ago

Kepler found that the orbits of the planets were _____. circular amorphous oval elliptical

vovangra [49]

<span>Kepler found that the orbits of the planets were elliptical.
His work was so convincing that to this day, they still are. </span>

7 0

2 years ago

Read 2 more answers

Consider the case in which the clay is launched along Path Y. The sphere of clay is launched with an initial velocity of v0 and

Helen [10]

Answer:

w = $\frac{m}{ml^2 + I R^2 }$ v l

Explanation:

Let's form a system formed by the clay sphere and the rod, in this case the angular momentum is conserved

initial instant. Before the crash

L₀ = m v l

Final moment. After the collision with the clay stuck to the rod

L_f = I_{total} w

angular momentum is conserved

L₀ = L_f

m v l = I_total w

w = $\frac{m}{I_{total} }$ v l

the total moment of inertia is the sum of the moments of inertia of the two bodies

the moment of inertia of the rod is

I_rod = I R²

I_total = m l² + IR²

we substitute

w = $\frac{m}{ml^2 + I R^2 }$ v l

4 0

2 years ago