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3 years ago

What is cubical expansivity of liquid while freezing

Physics

2 answers:

Elena L [17]3 years ago

7 0

Answer:

"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web

Explanation:

tbh up above ✅

kolezko [41]3 years ago

7 0

Answer:

cubic meter

Explanation:

Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion

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2.)

Oduvanchick [21]

Answer:

$-22.7 m/s^2$

Explanation:

This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem,

u = 27.8 m/s

v = 0

s = 17 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-27.8^2}{2(17)}=-22.7 m/s^2$

4 0

3 years ago

Which of the following is an example of a medium for a visible light wave?

goldfiish [28.3K]

Answer:

C. green light

Explanation:

4 0

3 years ago

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The Great Red Spot, a hurricane-like storm with twice the diameter of earth, is found on (a.)Jupiter (b.)Saturn (c.)Titan (d.)Th

Zielflug [23.3K]

Hello

The correct answer is A. Jupiter

hope this helps

plz mark me as brainliest

3 0

3 years ago

Read 2 more answers

A sealed beaker of what you are told is aqueous nickel sulfide was given to you by the local chemist and you put it on the windo

Masteriza [31]

Answer: D. It is a SUSPENSION

Explanation:

SUSPENSION

This is a combination of two or more single substances. The properties of the components involved are not however changed or lost as is the case with compounds.

For this reason this mixture can be separated due to sedimentation or filtering.

After a few days, this occurs in the aqueous nickel sulfide because the solid nickel sulfide is separating from the water.

7 0

3 years ago

Which of the following can penetrate the deepest (Please explain)

harkovskaia [24]

Answer: 3MeV electron

Explanation:

m_e={9.1\times 10^{-31} m_α=4\times m_e m_a={9.1\times 10^{-31}

m_p=1.67\times 10^{-27}

(a) K.E. Energy of electron = $\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}$ =3MeV

$v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }$ =1.05\times10^{18}

$v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}$

(b) K.E. Energy of alpha particle = $\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}$ =10MeV

$v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6} }{4\times9.1\times 10^{-31} }$ =0.88\times10^{18}

$v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}$

$v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }$ =0.035\times10^{18}

$v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}$

(d) K.E. Energy of proton particle = $\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}$ =400keV

$v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3} }{1.67\times 10^{-27} }$ =0.766\times10^{14}

$v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}$

from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.

6 0

3 years ago