What’s the ion of element iodine ?

How can the IMA of a first- class lever be increased?

Dimas [21]

IMA = Ideal Mechanical Advantage

First class lever = > F1 * x2 = F2 * x1

Where F1 is the force applied to beat F2. The distance from F1 and the pivot is x1 and the distance from F2 and the pivot is x2

=> F1/F2 = x1 /x2

IMA = F1/F2 = x1/x2

Now you can see the effects of changing F1, F2, x1 and x2.

If you decrease the lengt X1 between the applied effort (F1) and the pivot, IMA decreases.

If you increase the length X1 between the applied effort (F1) and the pivot, IMA increases.

If you decrease the applied effort (F1) and increase the distance between it and the pivot (X1) the new IMA may incrase or decrase depending on the ratio of the changes.

If you decrease the applied effort (F1) and decrease the distance between it and the pivot (X1) IMA will decrease.

Answer: Increase the length between the applied effort and the pivot.

4 0

4 years ago

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I need help. <br> How do I fill this out?

enyata [817]

Answer:

maybe try searching it up

6 0

3 years ago

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A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie

-BARSIC- [3]

a) $3.27\cdot 10^{-3} J$

b) $11.60\cdot 10^{-3} J$

c) $8.33\cdot 10^{-3} J$

Explanation:

a)

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

$C=13.5 \mu F = 13.5\cdot 10^{-6}F$ is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

$U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J$

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

$C'=kC$

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

$U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2$

where:

$C=13.5\cdot 10^{-6}F$ is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

$U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J$

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

$U=3.27\cdot 10^{-3} J$

The final energy stored in the capacitor, after the dielectric is inserted, is

$U'=11.60\cdot 10^{-3} J$

Therefore, the change in energy of the capacitor during the insertion is:

$\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J$

So, the energy of the capacitor has increased by $8.33\cdot 10^{-3} J$

8 0

3 years ago

What happens when exhale

polet [3.4K]

When you breathe in, or inhale, your diaphragm contracts (tightens) and moves downward. This increases the space in your chest cavity, into which your lungs expand. The intercostal muscles between your ribs also help enlarge the chest cavity. They contract to pull your rib cage both upward and outward when you inhale.

4 0

4 years ago

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A rectangular block measures 4.1cm by 2.8cm by 2.1cm. calculate its volume given you answer to an appropriate number of signific

sattari [20]

Answer:

Volume = 24 cm^3

Explanation:

We recall that the volume of the box is calculated via the formula:

Volume = length * height * width

and that in a product, the number of significant figures for the result should coincide with the number of significant figures of the factor that has the least of them.

In this case, all measures have the same number of significant figures: two. so we calculate the product, and then limit the answer value to exactly two significant figures:

Volume = 4.1 cm * 2.8 cm * 2.1 cm = 24.108 cm^3, which must be rounded to two significant figures as: 24 cm^3

6 0

3 years ago

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