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3 years ago

What does a step-up transformer do?

Physics

2 answers:

Lostsunrise [7]3 years ago

8 0

C.it steps up the voltage

Alla [95]3 years ago

4 0

Answer:

C. It steps up the voltage.

Explanation:

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What does this picture mean?

ivann1987 [24]

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be aware of large trucks on the roadway
Tip: usually, these signs are hazard signs or warning.
Also, this isnt physics </span><span />

5 0

3 years ago

Read 2 more answers

(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 $4\pi$ .

5 0

3 years ago

A wave is described by y=0.0200 sin (kx - ωt) , where , ω = 3.62 rad/s, x and y are in meters, and t is in seconds.(b) the wavel

BartSMP [9]

For a wave is described by y=0.0200 sin (kx - ωt) , where , ω = 3.62 rad/s, x and y are in meters, and t is in seconds, the wavelength = 2.978

<h3>How to solve for the wavelength</h3>

What is wave speed?

This is used to refer to the speed at which a wave is moving. It is the product of frequency and wave number

Given data

y=0.0200 sin (kx - ωt)

ω = 3.62 rad/s

y are in meters

t is in seconds

k = 2.11 rad/m

k = wavenumber = 2 * pi / wavelength

wavelength = 2 * pi / wavenumber

wavelength = 2 * pi / 2.11

wavelength = 2.978