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Lubov Fominskaja [6]

3 years ago

5

PLEASE HELP!!!!!!!!Move on to electric force. Blow up the two balloons and knot them. Then tie a thread onto each balloon. Suspe

nd the two balloons using tape so that they’re about six inches apart, and check that they don’t move or interact. Rub both balloons with wool or fur. If wool or fur is not available, rub the balloons on your hair. Do they attract or repel (push away) each other?

2 answers:

Triss [41]3 years ago

8 0

Answer:

they attract

Explanation:

timama [110]3 years ago

6 0

They attract. This is because of the Stagnant you have created.

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Speed that does not change is known as the:

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Answer: Final speed

Explaination: because its final.

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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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2 years ago

Prentice Hall. All

iVinArrow [24]

Answer:

Decreases/Reduces

Explanation:

Fill in the blank:

Consider the equation Work = Force X Distance.

<em>If a machine increases the distance over which a force is exerted, the force </em>

<em>required to do a given amount of work</em> .........

If the work is a constant value, then by isolating force from the equation, we get:

Force = Work / Distance

By increasing the value of the Distance, then the quotient Work. Distance diminishes, and therefore the required force decreases (diminishes, reduces)

Answer: Decreases/Reduces

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Colder in Alaska, warmer in Mexico.

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