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2 years ago

You have a 1.7 μf and a 2.2 μf capacitor. what values of capacitance could you get by connecting them in parallel?

Physics

2 answers:

Rus_ich [418]2 years ago

5 0

You have to add 1.7 and 2.2 together!
1.7+2.2=3.9μF

Dominik [7]2 years ago

5 0

The effective capacitance of capacitors in parallel is
the sum of the individual capacitances.

The only value of capacitance you could get with
those two in parallel is

(1.7 uF + 2.2 uF) = 3.9 uF

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2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a n

AleksandrR [38]

Answer:

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

$d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}$

For the ships to sight each other, distance must be 5 or smaller

$d \leq 5$

$\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5$

$(12 - 12t)^2 + (9t)^2 \leq 25$

$144t^2 - 288t + 144 + 81t^2 - 25 \leq 0$

$225t^2 - 288t + 119 \leq 0$

$(15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0$

$(15t^2 - 9.6)^2 + 26.84 \leq 0$

Since $(15t^2 - 9.6)^2 \geq 0$ then

$(15t^2 - 9.6)^2 + 26.84 > 0$

So our equation has no solution, the answer is no, the 2 ships never sight each other.

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3 years ago

In order to design an experiment, you need a ____ about the scientific question you are trying to answer.

goldenfox [79]

In order to design an experiment, you need a hypothesis about the scientific question you are trying to answer.

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3 years ago

Read 2 more answers

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

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3 years ago

If 35 mg of Oxygen is mixed with 1000 g of water at 25 degrees Celsius, a solution forms.

Serjik [45]

Answer:

The solute is oxygen

The solvent is water

Explanation:

A solvent is any chemical substance that dissolves other chemical substances, while a solute refers to any chemical substance that dissolves in other chemical substances. The best way to know when a chemical substance dissolves in another chemical substance is when a solid or gas dissolves in water. The solid or the gas can now be referred to as the solute and it will be shown to be in the the aqueous state, while the solvent is usually shown to be in the liquid state in any chemical equation.

Let us use the particular example of the dissolution of oxygen gas in water as shown below;

O2(g) + H2O(l)⇄O2(aq) + H2O(l)

The aqueous oxygen is the solute while the liquid water is the solvent.

Also, the substance having a smaller mass must be the solute and the substance having the larger mass must be the solvent.

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3 years ago

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siniylev [52]

We need to learn about median and mode because they are both important in statistical data and statistical data is what shows us important information about the world around us.

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