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The sodium ions and chloride ions are held together by the strong ELECTROSTATIC attractions between the POSITIVE and negative charges. You NEED one sodium ATOM to provide the EXTRA electron for one chlorine atom, so they combine together 1:1.

Here is the ANSWER EXPLANATION:PLZ MAKE BRAINLIEST answer

Dispersion of light occurs when white light is separated into its DIFFERENT constituent colors because of refraction and Snell's law. ... White light enters a PRISM on the LEFT, then is separated according to wavelength into a RAINBOW pattern.Explanation:

Given f=200cm u=100cmv=?mirror formula1/f=1/u+1/v by sign convention1/200=1/-100+1/v1/v=1/200+1/100=3/200v=200/3=66.6cmtherefore, the IMAGE is formed at 66.6cm BEHIND the mirror and it is virtual image .m=-v/u=-66.6/-100=66.6/100=0.66therefore , the image is erect and SMALLER in SIZE by a factor of 0.66Explanation:MARK ME BRILLIANT AND LIKE IT

tion:exactly the same volume as it displaced before. So the ADDED volume is the same, so the level of the WATER will not change. ... The water level remains the same when the ice cube melts. A FLOATING object displaces an amount of water equal to its own weight.hope it helps

In a TWO STROKE engine, the WORKING CYCLE is completed in two revolutions of the crankshaft. Explanation: In a two stroke engine, the working cycle is completed in one revolution of the crankshaft.Explanation:I hope it's HELP you..

:-❤We can CALCULATE the drift VELOCITY USING the EQUATION I = nqAvd. The current I = 20.0 A is given, and q = –1.60×10–19 C is the charge of an electron.

.005 kg20 cm = 0.2 mGravitational POTENTIAL Energy = mgh = (0.005)X(10)x(0.2) = 0.001 J

ans.c DISPLACEMENT TAKES PLACE

hoep it HELPS jExplanation:matter is made up of tiny PARTICLES CALLED ATOMS and MOLECULES.

A number of the FORM QP , where ′ P ′ and ′ Q ′ are any integers and q=0 is called a rational number. ∴ 01 is not a rational number because it is not defined.

We know that a non-luminous OBJECT is ONE that does not emit its own light, ALTHOUGH it can still reflect light from other sources. So, we can see a non-luminous object, when light is reflected from the object TOWARDS our eyes.

The main climate factors are direction of WIND , temperature and LOCATION .Explanation:Plz mark as BRAINLIEST...

The LOUDNESS of a NOTE is determined by its AMPLITUDE. (ii) The pitch of a note determined by its FREQUENCY. (iii) The quality of a note is determined by its wave form.Explanation:hope it's helpful

The great SCIENTIST Newton must be appreciated when it comes to the topic - SCATTERING OF LIGHT. He conducted an experiment which proved that when two IDENTICAL prisms are KEPT close to each other (one in inverted POSITION) , the ray of light passing through them CONVERGES.

5 KM east20km NORTH -10 km south = 10 km northdisp == 11.1 km APPROX

Hysteresis loops describe the magnetic response of a sample exposed to a magnetic field. The shape of a hysteresis loop can be diagnostic of the mineralogy of the sample (Channell and MCCABE, 1994; Tauxe, 1993). Ferrimagnetic minerals tend to produce narrow loops, whereas the presence of antiferromagnetic minerals causes hysteresis loops to be wider. Hysteresis parameters can also be used to estimate magnetic grain size (Day et al., 1977). When measuring a hysteresis loop, the sample is exposed to a maximum field Bmax and its magnetization (M) is measured as the field is reduced to zero, and increased again in the opposite direction all the way to the maximum Bmax. The REMANENCE measured at zero fields is equivalent to IRM. The field BC necessary to reduce the magnetization to zero is called the coercive force, and can be used to estimate the bulk coercivity of the sample. At high fields, the CURVE M(B) becomes linear and the increase in M is due to the presence of para- or diamagnetic minerals. This high field slope is known as high-field paramagnetic susceptibility. The saturation magnetization, MS, of a sample can be calculated from a hysteresis loop by extrapolating the linear high-field part of the loop back to B = 0. MS is a function of all magnetic minerals (ferri- and antiferromagnetic) present in a sample.Hysteresis loops can be measured either in a maximum field of 1250 mT, using a vibrating sample magnetometer (VSM), or in a maximum field of 5 T, using a Quantum Design MPMS-2 magnetic-properties measurement system, equipped with a superconducting magnet. These high fields are necessary to fully SATURATE SAMPLES. Coercive force (BC), SIRM, and saturation magnetization (MS) can be measured using Quantum Design MPMS-2 magnetic-properties measurement system.

consider two FACES of the cube, say P and Q, normal to x−axis.When the MOLECULE collides with the side Q with a velocity u, it rebounds with −u, while its other components remain unchanged.Change in momentum of the molecule due to this COLLISION =−2mu, which is imparted to the wall PER collision.The time taken by molecule to cover dis

I have to pay for the chapter 4 and I will be the PROCEDURE of the experiment solubility. The COUNTRY is country is divided into six states and is often LISTED as ONE of the largest countries on earth line between the Pacific and Road and is a house usually made of Brick Lane and is a house usually made of Brick Lane and is

tion:Carbon MONOXIDE is a highly poisonous, odorless, colorless, and tasteless GAS. It is very flammable in air over a wide range of concentrations (George 2001) and burns in air with a bright blue flame (O'Neil et al. 2006). It BECOMES a LIQUID at 81.62 K (-191.53 °C) and is insoluble in water above 70 °C (George 2001).

but where is figureshow me FIG 8.18

no the driving FREQUENCY of a paricular FORCE is not constantobviously it DEPENDS upon the mass of oscillator

Mass of the ball (m) = 200 gHeight (h) = 5 mAcceleration due to gravity (g) = 9.8 m/s²To Find :kinetic energy of the ball when it reaches the ground = ?Solution :First convert the unit of the mass of ball from gram to kilogram :→ Mass of ball = 200 g→ Mass of ball = (200) ÷ (1000)→ Mass of ball = 2 ÷ 10→ Mass of ball = 0.2 kgHence,the mass of the ball is 0.2 kg.According to the QUESTION Now :Finding potential energy at higest point :→ Potential energy = mass × Acceleration due to gravity × Height→ Potential energy = 0.2 × 9.8 × 5→ Potential energy = 1.96 × 5→ Potential energy = 9.8 JHence,the potential energy of the ball is 9.8 J.As we KNOW that, the Potential energy at the highest point is equal to the Kinetic energy at the lowest point.Hence,from the above STATEMENT we can CONCLUDE that the Kinetic energy of the ball when it reaches the ground will be 9.8 J.

The velocity vector is constant in magnitude but CHANGING in DIRECTION. Because the speed is constant for such a motion, many STUDENTS have the MISCONCEPTION that there is no acceleration. ... For this reason, it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating.Explanation:hope it's HELPFUL

1. YUNIT II ARALIN 2 PUNA AT MUNGKAHI MO, TANGGAP KO LAYUNIN Nakapagpapakita ng pagkamahinahon sa damdamin at KILOS ng kapuwa tulad ng pagtanggap ng puna ng kapuwa nang maluwag sa kalooban2. Walang perpektong tao sa mundong ito. Lahat TAYO ay MAY kahinaan at kakayahan, subalit magkakaiba ang pagtanggap ng tao sa kanilang mga3. Day 1 Alamin Natinplease make me brianllest answer

D. MLT−1Q−1Explanation:PLS MARK as BRAINLIEST

ferences between LIGHT and soundare as follows: ... Light WAVES areelectromagnetic waves while sound wavesare MECHANICAL waves. Light waves are transverse while sound waves are longitudinal. Light waves can TRAVEL in vacuum.

্যান্ডসাউন্ডের মধ্যে পার্থক্যগুলি নিম্নরূপ: ... হালকা তরঙ্গগুলি বৈদ্যুতিন চৌম্বক তরঙ্গগুলি যখন সাউন্ড তরঙ্গগুলি যান্ত্রিক তরঙ্গ IGHT হালকা তরঙ্গগুলি ট্রান্সভার্স হয় তবে সাউন্ড তরঙ্গগুলি দ্রাঘিমাংশ হয়। লাইট

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StateThe state is a FORM of HUMAN ASSOCIATION distinguished from other SOCIAL groups by its purpose, the establishment of order and security; its methods, the laws and their enforcement; its territory, the area of jurisdiction or geographic BOUNDARIES; and finally by its sovereignty.

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InvertedExplanation:REAL IMAGES (images on the...” ... Real images (images on the same SIDE of the OBJECT) are always inverted. Virtual images (images on OPPOSITE side of an object) are always erect/ upright.

es are the PEOPLE and their work that are INVOLVED in PROVIDING a PARTICULAR service.Hope it helps you.

Explanation :Here, initally P Here, initally P 1Here, initally P 1 Here, initally P 1 = PHere, initally P 1 = PV Here, initally P 1 = PV 1Here, initally P 1 = PV 1 Here, initally P 1 = PV 1 =V+V = 2V;Here, initally P 1 = PV 1 =V+V = 2V;Finally, P Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VHere, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as INFERRED from equation)Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VHere, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VP Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VP 1Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VP 1 Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VP 1 V Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VP 1 V 1Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VP 1 V 1VHere, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VP 1 V 1VP×2V Here, initally P 1 = PV 1 =V+V = 2V;Finally, P 2 = P; V 2 = VAS P 1 V 1 = P 2 V 2 (as inferred from equation)P 2 = VP 1 V 1VP×2V =2P