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3 years ago

Someone help this is a test and i need to finish this quick i will give brainliest of it lets me

Physics

1 answer:

tekilochka [14]3 years ago

7 0

Sorry, I don't know but I think the correct answer is the first option.

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Need help quick!!!!

BARSIC [14]

The answer is B. Unbalanced force

3 0

3 years ago

Read 2 more answers

If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no

Yakvenalex [24]

<u>Answer</u>

3.7 Km south

<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

(+1.7) + (-5.4) = 1.7 - 5.4

= - 3.7 Km But negative was towards south.

∴ Answer = 3.7 Km south.

6 0

3 years ago

Which term is applied to an object through which light passes?

Elena L [17]

Answer:

Explanation:

transparent_objects that allows light to pass through and can you see through them

5 0

3 years ago

Read 2 more answers

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, the

ICE Princess25 [194]

Complete Image

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by $\Delta L_{thick}$ and $\Delta L_{thin}$ , respectively.The ratio ΔLthick rod/ΔLthin rod is:

a.) =1

b.) <1

c.) >1

Answer:

The ratio is less than 1 i.e $\frac{1}{2}$ option B is correct

Explanation:

The Young Modulus of a material is generally calculated with this formula

$E = \frac{\sigma}{\epsilon}$

Where $\sigma$ is the stress = $\frac{Force}{Area}$

$\epsilon$ is the strain = $\frac{\Delta L}{L}$

Making Strain the subject

$\epsilon = \frac{\sigma}{E}$

now in this question we are that the same tension was applied to both wires so

$\frac{\sigma}{E}$ would be constant

Hence

$\frac{\Delta L}{L} = constant$

for the two wire we have that

$\frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2}$

Looking at young modulus formula

$E = \frac{\frac{F}{A} }{\frac{\Delta L}{L} }$

$E * \frac{\Delta L }{L} = \frac{F}{A}$

$A * \frac{\Delta L}{L} = \frac{F}{E}$

Now we are told that a comprehensive force is applied to the wire so for this question

$\frac{F}{E}$ is constant

And given that the length are the same

$A_1 \frac{\Delta L_{thin}}{L_{thin}} = A_2 \frac{\Delta L_{thick}}{L_{thick}}$

Now we are told that one is that one rod is twice as thick as the other

So it implies that one would have an area that would be two times of the other

Assuming that

$A_2 = 2 A_1$

$A_1 \frac{\Delta L_{thin}}{L_{thin}} = 2 A_1 \frac{\Delta L_{thick}}{L_{thick}}$

$\frac{\Delta L_{thin}}{L_{thin}} = 2 \frac{\Delta L_{thick}}{L_{thick}}$

From the question the length are equal

$\Delta L_{thin} =2 \Delta L_{thick}$

$\frac{\Delta L_{thick}}{\Delta L_{thin} } = \frac{1}{2}$

Hence the ratio is less than 1

6 0

3 years ago

A switch that connects a battery to a 30μf capacitor is closed. several seconds later you find that the capacitor plates are cha

Elanso [62]

You can find the emf of the battery by using:

Q = CV.

Q = Charge in capacitor plates = 30μC
C = Capacitance of the capacitor = 30μF

V = Q/C = 30/30 = 1V

Ans: The emf of the battery = 1V.

5 0

3 years ago