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3 years ago

Calculate how much work is done if a current of 2A flows through a potential difference of 12V for 2 minutes

Physics

1 answer:

Sedaia [141]3 years ago

3 0

W= P . t

P= I . V

= 2. 12

= 24 watt

then,

W= 24 . 2.60 (convert to second)

= 2880 J

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One’s behavior cannot be affected by one’s subconscious

igomit [66]

That statement is false

Our subconscious tend to give a strong influence to our behaviours even if we are not feeling it directly.

For example, let's say that there is a boy that hurt by cats and it's ingrained in his head that cats possess high level of danger to him. Even after that boy grow up, his subsconcious would most likely cause a certain level of paranoia that make him either scared of cats or simply annoyed by seeing them.

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3 years ago

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(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

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3 years ago

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timofeeve [1]

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Find the speed of a wave with wave length 650 M and a frequency 35HZ<br><br> help.

hram777 [196]

Answer:

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How do you solve <img src="https://tex.z-dn.net/?f=4x%5E%7B3%7D" id="TexFormula1" title="4x^{3}" alt="4x^{3}" align="absmiddle"

Vlad1618 [11]

$Hello$ $There!$

Here's a explanation!

Let's solve your equation step-by-step.

$4x^3=2x^-^1$

$4x^3=\frac{2}{x}$

Step 1: Multiply both sides by x.

$4x^4=2$

$\frac{4x^4}{4} =\frac{2}{4}$

(Divide both sides by 4).

$x^4=\frac{1}{2}$

$x=+(\frac{1}{2} )^(^\frac{1}{4} ^)$

Take the root.

ANSWER!

$x=0.840896$ $Or$ $x=-0.840896$

Hopefully, this helps you!!

$AnimeVines$

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