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2 years ago

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When a 15.00 kg mass is attached to a vertical spring, the spring is stretched 2.0 m such that the mass is 6.0 m above the table

. The spring constant is 400.0N/m? What is the total potential energy associated with this system?

1 answer:

Marysya12 [62]2 years ago

3 0

Answer:

idk

Explanation:

idk

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When using science to investigate physical phenomena, which characteristic of the event must exist?

Vesna [10]

The correct answer is B. Measurable

Explanation:

The use of science to investigate a phenomenon implies using measurements and observations to better understand a phenomenon or test a hypothesis. Moreover, science focuses on natural phenomena that can be objectively studied through measurement instruments such as a thermometer, balance, hydrometer, etc.

In this context, for a phenomenon to be studied by science this needs to be measurable because the use of precise instruments as wells as numbers allow scientist to analyze and understand a phenomenon. Moreover, phenomena that depend on personal perspectives and cannot be measure is considered as non-scientific.

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3 years ago

The potential difference between a pair of oppositely charged parallel plates is 398 V. If the spacing between the plates is dou

LiRa [457]

Answer:

Explanation:

capacitance of parallel plate capacitor

c = ε A / d , d is distance between plates , A is surface area , ε is constant

As d becomes two times , Capacitance c = 1/ 2 times ie c / 2

potential V = Q / C

Q is constant , potential

v = Q / c /2

= 2 . Q / C

= 2 V

So potential difference becomes 2 times.

NEW P D = 398 X 2

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3 years ago

yarga [219]

The answer would be C

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3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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3 years ago

Newton’s second law states that the acceleration of an object is dependent on 1 variable which is the net force

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