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2 years ago

As the cross-sectional area of a conductor increases, its resistance .

Physics

2 answers:

n200080 [17]2 years ago

8 0

Answer:

Yess it’s resistance

Explanation:

Reptile [31]2 years ago

8 0

Answer:

DECRAAAASESSSS

Explanation:

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Find the resistance offered by the nail A.

Nana76 [90]

3.4 × 10³ N.

<h3>Explanation</h3>

The perpendicular distance between the 400 N force and the fulcrum is $120 \; \text{mm} = 0.120 \; \text{m}$ .

What's the perpendicular distance between the friction of the nail and the fulcrum?

Refer to the diagram attached. The perpendicular distance between the nail and the fulcrum is $0.015 \; \cos{20\textdegree{}} = 0.0141 \; \text{m}$ .

What's the size of the friction of the nail?

$F_\text{friction} \cdot l_\text{friction}= F_{\text{400} \; \text{N}} \cdot l_{\text{400} \; \text{N}}$

$F_\text{friction} = \dfrac{F_{\text{400} \; \text{N}} \cdot l_{\text{400} \; \text{N}}}{ l_\text{friction}}\\\phantom{F_\text{friction}} = \dfrac{400\times 0.120}{0.0141}\\\phantom{F_\text{friction}} = 3.4\times 10^{3} \; \text{N}$ .

8 0

2 years ago

You have a 100 g sample of Unobtainium, it is half a life of 2 years. How much will be left after 10 years?

sergeinik [125]

3.125 ,, that is the answer !!

3 0

2 years ago

A study of how much corn was produced in a field depending on the planting density compared the yield from fields planted with 4

MariettaO [177]

Answer:

$\frac{d_2}{d_1} =\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}= 1.225$

Or in the other way:

$\frac{d_1}{d_2} =\frac{\sqrt{4}}{\sqrt{6}}=\sqrt{\frac{4}{6}} = \frac{\sqrt{6}}{3}= 0.816$

Explanation:

For this case we need to remember that hectare is a measure of area. And we have the following ratios:

$r_1 =40000 \frac{plants}{acre}$

We can convert this into $plants/m^2$

$r_1 = 40000 \frac{plants}{acre} *\frac{1 acre}{10000 m^2}=4 \frac{plants}{m^2}$

$r_2=60000 \frac{plants}{acre}$

We can convert this into $plants/m^2$

$r_2 = 60000 \frac{plants}{acre} *\frac{1 acre}{10000 m^2}=6 \frac{plants}{m^2}$

By dimensional analysis we assume that $A = L^2$ we can find the ratio in terms of distance like this:

$d_1 =\sqrt{4}= 2 \frac{\sqrt{plants}}{m}$

$d_2= \sqrt{6}= 2.449 \frac{\sqrt{plants}}{m}$

And if we find the ratios in terms of distance we got:

$\frac{d_2}{d_1} =\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}= 1.225$

Or in the other way:

$\frac{d_1}{d_2} =\frac{\sqrt{4}}{\sqrt{6}}=\sqrt{\frac{4}{6}} = \frac{\sqrt{6}}{3}= 0.816$

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2 years ago

the modulus of the vector product of two vector is 0.577 times their scalar product the angle between vectors

BaLLatris [955]

What exactly do u want me to do for u mam/sir

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2 years ago

An example of kinetic energy is ______.

ivanzaharov [21]

Kinetic energy is the energy of motion, and things that are
in motion have it. The only choice on the list that mentions
anything moving is the "car coming to a stop".

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2 years ago

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