What is shown in the above figure

Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the

Nata [24]

<u>Explanation</u>:

For series

$\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)$

$$I=I_{1}=I_{2}=I_{n}$ (current is the same)$

$V=I R(\text {voltage is directly proportional to } R)$

$R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }$

For parallel

$\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })$

$I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})$

$I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)$

$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}$

3 0

3 years ago

How do you know which forces works for free bodies

miss Akunina [59]

Answer:

Gravitational force (pulled downward by the Earth)

Normal force (pushed upward by the ground)

Applied force (pushed by the person)

Friction force (pulled opposite the direction of motion by the roughness of the ground)

5 0

2 years ago

I don’t understand this

blondinia [14]

Answer:

Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of $160cm^{2}$ by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.

Have an amazing day and I hope this can somewhat help :)

7 0

2 years ago

In a major human artery with an internal diameter of 5mm, the flow of blood, averaged over the cardiac cycle is 5cm3·s−1. The ar

antiseptic1488 [7]

9514 1404 393

Answer:

see attached

Explanation:

Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.

(5 cm³/s)(1.06 g/cm³) = 5.3 g/s

If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:

(5.3 g/s)/2 = 2.65 g/s

__

The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...

(5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s

Downstream, we have half the volumetric flow and a smaller area.

(2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s

7 0

2 years ago

A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The channel undergoes a smooth, gradu

Trava [24]

Answer:

Depth in the contracted section = 2.896m

Velocity in the contracted section = 2.072m/s

Explanation:

Please see that attachment for the solving.

Assumptions:

1. Negligible head losses

2. Horizontal channel bottom

7 0

2 years ago