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3 years ago

What technology produced the results illustrated here, and what is this technology used for?

Physics

2 answers:

IrinaVladis [17]3 years ago

6 0

Where is the phone ? Can’t answer without a picture of the problem

elixir [45]3 years ago

4 0

The results illustrated here were hidden by means of Obscurational Cloaking technology.

This technology is used to hide, conceal, obscure, and withhold results.

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What type of image is formed by a lens if m = -2.0?

natka813 [3]

The best answer seems to be the C) <span>A virtual image that is smaller than the object. Because when m is smaller than 1, then the image is virtual. But I'm not 100% sure.</span>

7 0

3 years ago

Two identical point charges in outer space are held apart at a distance D. As soon as the charges are released, each begins movi

AVprozaik [17]

Answer:

B. 4a

Explanation:

Force between the charges is inversely proportional to the square of the distance

=> Force will be 4 times and acceleration will be 4a

=> Answer b).

6 0

3 years ago

Read 2 more answers

Explain why pumice floats and why ironwood sinks

almond37 [142]

Answer:

A quick way of describing density is to describe an object as heavy or light for its size. Pumice stone, unlike regular rock, does not sink in water because it has a low density. An ironwood branch is very dense and sinks in water.

Hope that helps. x

6 0

2 years ago

Read 2 more answers

You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro

ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

$F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N$

The spring obeys Hook's law:

$F=k\Delta x$

where k is the spring constant and $\Delta x$ is the stretching of the spring. Since we know $\Delta x=2.92 cm=0.0292 m$ , we can re-arrange the equation to find the spring constant:

$k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m$

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

$F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N$

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

$\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm$

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance $\Delta x$ is equal to the elastic potential energy stored by the spring, given by:

$W=U=\frac{1}{2}k\Delta x^2$