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2 years ago

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of __________.

Physics

1 answer:

Dahasolnce [82]2 years ago

7 0

Answer: young's modulus

Explanation: A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of young's modulus .

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PLEASE HELP!!! DUE SOON!!! NO LINKS OR RANDOM WORDS!!

Katyanochek1 [597]

Increase to the visible part

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3 years ago

A 67.5-kg person throws a 0.0410-kg snowball forward with a ground speed of 34.0 m/s. A second person, with a mass of 57.5 kg, c

lora16 [44]

Answer:

Final velocity of the first person is 3.43m/s and that of the second person is 0.0242m/s

Explanation:

Let the momentum of the first person, the ball second person be Ma, Mb and Mc.

From the principle of the conservation of momentum, sum of the momentum before collision is equal to the sum of the momentum after collision.

Ma1 + Mb1 = Ma2 + Mb2.

The ball and the first person are both moving together with a common velocity 3.45m/s.

Let the velocity of the first person be v1

Therefore

67.5×3.45+ 0.041×3.45= 67.5v1 + 0.041×34

233.02 = 1.39+ 67.5v1

67.5v1 = 233.02 - 1.39 = 231.61

v1 = 231.61 / 67.5

v1 = 3.43m/s

The second person and the ball move together with a common velocity after catching the ball.

For the second person and the ball let their final common velocity be v

Mb2 + Mc2 = Mb3 + Mc3

0.041 × 34 + 57.5 ×0 = (57.5 + 0.041)×v

57.541v = 1.39

v = 1.39 /57.541

v = 0.0242m/s

5 0

4 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Phoenix [80]

Answer:

The temperature of the core raises by $2.8^{o}C$ every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is $1.60\times 10^{5}kg$ will require

$0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ$

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

$\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s$

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3 years ago

Find the work done "by" the electric field on a positively charged point particle with a charge of 2.1x10^-6 C as it is moved fr

Alexxx [7]

Answer

Work done will be $14.7\times 10^{-6}J$ and it will be positive

Explanation:

We have given charge $2.1\times 10^{-6}C$

We have to find work done in moving the charge from 15 volt to 8 volt

Let $V_1=15V\ and\ V_2=8volt$

So potential difference $V=V_1-V_2=15-8=7volt$

We know that work done $W=QV$ , here Q is charge and V is potential difference

So work done $W=QV=2.1\times 10^{-6}\times 7=14.7\times 10^{-6}J$

It will be positive work done because work is done in moving charge from higher potential to lower potential

5 0

4 years ago

Which statements accurately describe matter

vladimir1956 [14]

<h3>I hope it is helpful for you ...</h3>

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3 years ago