Ask question

Ask question

All categories

3 years ago

13

Which statement describes one feature of a closed circuit? Charges do not flow. Bulbs will not shine. The circuit is broken. The

circuit is complete.

2 answers:

Novay_Z [31]3 years ago

7 0

I inferred you've referring to a close electrical circuit.

Answer:

The circuit is complete.

Explanation:

A closed electrical circuit is indeed a complete circuit. Also, it allows charges to flow, the bulbs in the circuit will shine and it is not broken.

It is termed closed circuit because there is no brokage in the series of electrical wires or the switch; which may prevent the free flow of current or charges. Thus, a feature that marks closed circuits is that they are complete.

san4es73 [151]3 years ago

4 0

Answer:

The circuit is complete.

Explanation:

You might be interested in

.Which uplift mechanism is pictured here?

zhenek [66]

The correct answer is orographic uplift! Hope this helps

4 0

4 years ago

Read 2 more answers

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

B)

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

C)

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0

3 years ago

Four uniform spheres, with masses mA = 200 kg, mB = 250 kg, mC = 1700 kg, and mD = 100 kg, have (x, y) coordinates of (0, 50 cm)

rusak2 [61]

Answer:

$F_2=2.43\times10^{-19}\ N$

$\theta=21.496^{\circ}$ in the anticlockwise direction from the positive x- axis.

Explanation:

Given are the masses of various spheres with their names in subscript:

$m_A=200\ kg$

$m_B=250\ kg$

$m_C=1700\ kg$

$m_D=100\ kg$

Now their respective coordinate positions (in cm) are as given below:

$P_A=(0,50)$

$P_B=(0,0)$

$P_C=(-80,0)$

$P_D=(40,0)$

<u>Now force on B due to A:</u>

$F_{BA}=G\times \frac{200\times 250}{50^2}$

$F_{BA}=20G\ N$

<u>Now force on B due to C:</u>

$F_{BC}=G\times \frac{1700\times 250}{80^2}$

$F_{BC}=66.406G\ N$

<u>Now force on B due to D:</u>

$F_{BD}=G\times \frac{100\times 250}{40^2}$

$F_{BD}=15.625G\ N$

<em>We observe that the forces due to masses C&D act opposite in direction.</em>

<u>So, the net force in the x-direction:</u>

$F_x=F_{BC}-F_{BD}$

$F_x=66.406G-15.625G$

$F_x=50.781G\ N$ in the positive x-direction

<em>We have only one force in y-direction due to mass A.</em>

So,

$F_y=20G\ N$ in the positive y-direction.

<u>Now the net force:</u>

$F_2=\sqrt{F_y^2+F_x^2}$

$F_2=\sqrt{(20G)^2+(50.781G)^2}$

$F_2=54.5776G^2\ N$

$F_2=2.43\times10^{-19}\ N$

<u>Now the direction of this force with respect to x-axis:</u>

$tan\ \theta=\frac{F_y}{F_x}$

$tan\ \theta=\frac{20G}{50.781G}$

$\theta=21.496^{\circ}$ in the anticlockwise direction from the positive x- axis.

3 0

3 years ago

Read 2 more answers

A 3.75 kg ball is lifted from the floor to a height of 1.5 m above the floor. What is its increase in potential energy?

inysia [295]

<span>We can use a simple equation to calculate the increase in gravitational potential energy. PE = mgh m is the mass of the object g is the acceleration due to gravity h is the change in height PE = mgh PE = (3.75 kg) (9.80 m/s^2) (1.5 m) PE = 55.1 Joules The increase in gravitational potential energy is 55.1 Joules.</span>

4 0

3 years ago

Four monitoring wells have been placed around a leaking underground storage tank. The wells are located at the corners of a 1-ha

Readme [11.4K]

Answer:

direction : West to East

magnitude : 6.0 * 10^-3

Explanation:

<em>Given data :</em>

Four ( 4 ) monitoring wells

location of wells = corners of 1-ha square

Total piezometric head in each well ;

NE corner = 30.0 m ;

SE corner = 30.0 m;

SW corner = 30.6 m;

NW corner = 30.6 m.

<u>Calculate for the magnitude and direction of the hydraulic gradient </u>

first step ; calculate for area

Area = ( 1 -ha ) ( 10^4 m^2/ha )

= 1 * 10^4 m^2

Distance between the wells = length of side

= √( 1 * 10^4 ) m^2

= 100 m

Direction of Hydraulic gradient is from west to east because the total piezometric head in the west = Total piezometric head in the east

Next determine The magnitude of the hydraulic gradient

= ( 30.6 - 30 ) / 100

= 6.0 * 10^-3

<u />

<u />

8 0

3 years ago