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3 years ago

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An object is sitting on the floor. A 22.4 N force is pulling the object to the right and an 11 N force is pulling the object to

the left. The gravitational force on the object is 70 N. What is the net force?

1 answer:

Nimfa-mama [501]3 years ago

5 0

answer = 33.4 net force.

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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

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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

The power of a purely resistive lead is always positive although the current and voltage are sometimes negative. explain

pickupchik [31]

Answer:

Current is in phase with voltage in a resistive circuit. Note that the wave form for power is always positive, never negative for this resistive circuit. This means that power is always being dissipated by the resistive load, and never returned to the source as it is with reactive loads.Explanation:

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Sunburn is caused by ultraviolet light waves having a frequency of around 10^16 Hz. What is their wavelength if the speed of lig

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$Given:\\f=10^{16}Hz\\c=3\cdot 10^8\frac{m}{s} \\\\Find:\\\lambda =?\\\\Solution:\\\\\lambda = \frac{c}{f} \\\\\lambda= \frac{3\cdot 10^8\frac{m}{s}}{10^{16}Hz} =3\cdot 10^{-8}m = 30nm$

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A 50-kg toboggan is coasting on level snow. As it passes beneath a bridge, a 20-kg parcel is dropped straight down and lands in

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Answer:

$KE2/KE1=0.71$

Explanation:

By conservation of the linear momentum:

m1*V1 = (m1+m2)*V2

Solving for V2:

$V2 = \frac{m1}{m1+m2}*V1$

The kinetic energies are:

$KE1=1/2*m1*V1^2$

$KE2 = 1/2*(m1+m2)*V2^2$

$KE2 = 1/2*(m1+m2)*(\frac{m1}{m1+m2}*V1)^2$

Simplifying:

$KE2 = 1/2*\frac{m1^2}{m1+m2}*V1^2$

The ratio will be:

$KE2/KE1=\frac{1/2*\frac{m1^2}{m1+m2}*V1^2}{1/2*m1*V1^2} =\frac{m1}{m1+m2}$

$KE2/KE1=0.71$

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A television set is plugged into a 120 V outlet. The television circuit carries a current equal to 0.75 A. What is the overall r

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Answer:

20,00

Explanation:

This is the answer

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2 years ago