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3 years ago

What are 3 common sources of voltage difference?

1 answer:

3 0

We are asked to enumerate 3 different sources of potential difference and the answers are listed below:
1. There was two parallel sources and voltage of each source is different from each other.
2. Different batteries are connected.
3, Different values of resistors were used in the circuit.

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The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.

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The resistance is 27.5 ohms

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3 years ago

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A heavier car is always safer in a crash than a lighter car.

kvv77 [185]

Answer:

not true because the mass from the heavy car will cause it to damage more

Explanation:

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3 years ago

Select the correct answer from each drop-down menu.

kirill115 [55]

Answer:

Explanation:

I think but I am sure

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3 years ago

I need help on ideas for a science project.. (highschool ideas)

Nana76 [90]

A research question that would complete the third question you need that are related to the first 2 questions which are:

“what type of masks help prevent fog on glasses when breathing?”
“does a mask’s material affect the level of fog on glasses as an effect of breathing?”

Would be: "Are there any available masks that could prevent fog on glasses that could be improved upon"?

This new research question would help you find out if there is an already existing mask that could be made better.

<h3>What is a Research Question?</h3>

This refers to "a question that a research project sets out to answer". and seeks to give answers to particular phenomena.

Hence, we can see that the new research question Would be: "Are there any available masks that could prevent fog on glasses that could be improved upon"?

This new research question would help you find out if there is an already existing mask that could be made better.

Read more about research questions here:

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1 year ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

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

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

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

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

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

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

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3 years ago