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2 years ago

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The focal length of david's lens is 60 cm. if rebecca stands in front of david at a distance of do and david perceives the posit

ion of rebecca at di, what does do equal if the magnification is -0.40?

1 answer:

Makovka662 [10]2 years ago

5 0

if rebecca stands in front of david at a distance of do and david perceives the position of rebecca at di, di will be +84 cm

<h3>What is focal length ?</h3>

How strongly light converges or diverges depends on an optical system's focal length, which is the inverse of optical power. A system with a positive focus length is said to converge light, whereas one with a negative focal length is said to diverge light.

focal length = +60 cm

magnification m = -0.40

focal length being positive an magnification negative.

given lens is a convex lens.

for a lens

m = di/do and 1/f = (1/di) - (1/do)di

= -0.4do1/f = (1/-0.4do) - 1/do do

= -210 cmdi = -0.4 * -210

di = +84 cm

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The force of friction is 300 N

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force acting on the object

m is its mass

a is its acceleration

For the cart in this problem, we have two forces acting on it:

- The force of push, F = 500 N, forward

- The force of friction, $F_f$ , backward

So Newton's second law can be rewritten as

$F-F_f = ma$

where

m = 50 kg

$a=4 m/s^2$ is the acceleration of the cart

And solving for $F_f$ , we find the force of friction:

$F_f = F-ma=500-(50)(4)=300 N$

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3 years ago

A small steel wire of diameter 1.0 mm is connected to an oscillator and is under a tension of 7.5 N. The frequency of the oscill

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Answer:

a

The output power is $P= 0.764Watt$

b

The Amplitude would decrease by $\frac{1}{2}$

Explanation:

From the question we are told that

The diameter of the steel wire is = 1.0mm $= \frac{1}{1000} = 1.0*10^{-3}m$

The raduis of this steel wire is $r = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Now from the question we can deduce that the power output is equal to the power being transmitted by wave on the wire this is mathematically represented as

$P = \frac{1}{2} \mu w^2 A^2 v ----(1)$

Where $\mu$ is the mass per unit length of the wire

This is mathematically evaluated as

$\mu = a* \rho$

Where a is the area of the the wire = $\pi r^2 = (3.142 * 0.5*10^{-3})^2 =7.855*10^{-7}m^2$

$\rho$ is the density of steel with a generally value of $7850 kg/m^3$

So

$\mu = 7.855*10^{-7} *7850$

$= 6.162*10^{-3}kg/m$

$w$ is velocity of the wave

This is mathematically evaluated as

$w=2 \pi f$

substituting 60Hz for f

We have

$w = 2 *3.142 * 60$

$=377.04 \ rad/s$

$A$ is the amplitude with a given value of 0.50 cm $= \frac{0.50}{100} = 0.50 *10^{-2}m$

v is the linear velocity of the wave

This is mathematically evaluated as

$v = \sqrt{\frac{T}{\mu} }$

Where T is the tension with a given value of $7.5N$

$v = \sqrt{\frac{7.5}{6.162*10^{-3}} }$

$=34.89 m/s$

Substituting values into equation 1

$P = 6.162*10^{-3}* 377.04^2 * (0.5*10^{-2})^2 * 34.89$

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Since the doubling of the frequency does not affect the amplitude and from equation one the output power is $\ \frac{1}{2}$ of the Amplitude, Then the Amplitude would decrease by $\frac{1}{2}$

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<h3>What is pressure</h3>

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In this way, for the same force, the pressure is inversely proportional to the surface: this means that, for the same force, the smaller the surface, the greater the pressure exerted and the more the body will be deformed.

That is, pressure is defined as the force exerted perpendicularly on a surface and is expressed mathematically as:

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Then, the definition of force can be expressed as:

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