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Genrish500 [490]
3 years ago
15

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

Substituting k=907.5 N/m and \Delta x=8.80 cm=0.088 m, we find the amount of work that must be done:

W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J

5 0
3 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
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