This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What is the compound next to Butane in its homologous series. |
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Answer» The homologous series of straight-chained alkanes begins methane (CH4), ethane (C2H6), propane (C3H8), BUTANE (C4H10), and pentane (C5H12). In that series, successive members differ in MASS by an extra methylene bridge (-CH2- unit) INSERTED in the chain. |
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| 2. |
कोई केस में वह इस पकड़ लेंगे अनावश्यक ऑफ हाइट 1 सेंटीमीटर इज द परपेंडिकुलर टू द प्रिंसिपल फॉर द प्रिपरेशन ऑफ चैप्टर ऑफ इमेज फ्रॉम |
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Answer» ऑस्ट्रेलिया इंग्लैंड इंडिया पाकिस्तान बांग्लादेश |
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| 3. |
How do we see nearby objects clearly and distinctly?tel |
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Answer» with the HELP of MAGNIFYING glass we can SEE the object clearly |
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| 4. |
Q.3. A hammer of mass 500 g, moving at 50 m 5-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer ? |
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Answer» the FORCE is DIRECTLY PROPORTIONAL to the LENGTH of your LODA |
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| 5. |
Question 1 A F+ 41A disk A of radiusr moving on a perfectly smooth surface at a speed v undergoes an elastic collision with an identicalstationary disk B. Find the velocity of the disk A after a collision if the impact parameter is 1/2 as shown in the figure.DB |
| Answer» CHECK the below√4v V4 V2 3–√v2 ANSWER : A | |
| 6. |
129 महेश फोकल लेंथ ऑफ 10 सेंटीमीटर हुआ कैन वे अप्लाई साउथ साउथ एक्टर इमेज फोन स्मार्ट प्राइस |
| Answer» TION:POWER of lens = -2.0 DioptreLet the focal length of the lens be f.We KNOW that\begin{lgathered}Power \: of \: lens = \FRAC{1}{focal \: length} \\ \\ - 2.0 = \frac{1}{f} \\ \\ f = \frac{1}{ - 2}\end{lgathered} Poweroflens= focallength1 −2.0= f1 f= −21 f = -0.5 metreSince the focal length of the lens is negative. So, it is a concave lens. | |
| 7. |
a particle is revolving in a circular path of radius 1÷π m it distance after completing two and half rotation |
| Answer» HEY GUYS here is your ANSWER PLEASE FOLLOW me | |
| 8. |
A uniform chain of mass m hangs from a light pulley, with unequal lengths hanging from the two sides ofthe pulley. The force exerted by the moving chain onthe pulley is |
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| 9. |
Pls answer this as soon as possible..... |
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Answer» Initial velocity of CAR, u = 108km/h 108 * 1000/3600 108*10/36 3*10 30m/sNow, From position-velocity relation, v^2 - u^2 = 2as (30)^2 - 0 = 2*a*150 900/2 = 150a 450 = 150a a = 450/150 a = 3m/s^2OPTION C IS THE RIGHT ANSWER HOPE IT HELPS PLZ DO MARK ME AS THE BRAINLIEST I REALLY NEED IT |
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| 10. |
- Perform an activity to show that gravitational force is a non- Contactfonce and write it in the copy withsuitable pictures. [glass and coin) |
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Answer» tion:it was a good DAY at WORK and I don't KNOW what I WANT to be a good day at work |
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| 11. |
a particle is revolving in a circular path of radius 1:π m it distance after completing two and half rotation |
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Answer» A particle moves along a CIRCULAR path of 6m RADIUS such that the arc of the ... DISTANCE = 1/ 2 Circumference ... He has completed two and half rounds after STARTING from the corner |
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| 12. |
28. The given Wheatstone bridge is balanced when p = 100 22; Q = 1 k22 and S = 120 12. Calculate the valueof the unknown resistance.(a) 1.1 k 2(b) 1.2 ks2(c) 1.3 k2(d) 1.5 k2 |
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Answer» +₹++₹+₹(3+3(3+2+2 |
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| 13. |
For cooking all come from some source. We need to know howwe select the source needed for obtaining the energy in its usable forActivity 14.2alConsider the various options we have when we choose a fidorcooking our foodWhat are the enteria you would consider when ying to alegresomething as a good fuel?Would your choice be different if you livedfa) in a forestin a remote mountain village or small islandel in New Delhiid) red five centuries agoHow are the factors different in each case?After going through the two activities above, we can see that usource of energy. or fuel, we select for performing some o |
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Answer» Various fuel options for cooking food for us are:Electricity, LPG, WOOD, Coal, solar cooker, etc.In the home we prefer LPG and electricity as they are highly efficient, does not cause smoke, and are easy to use.Good fuel is a fuel which is easily available, cheap, clean, and easy to use.Activity 14.2A solar cooker is easy to use, cheap and clean source of energy to cook food.Criteria to CLASSIFY fuel:We classify fuels based on many criteria like;State of fuel: Solid, liquid, gas.Ease of use: Easy to handle, hard to handle. Here electricity and LPG are easy to handle while using wood is cumbersome.Economy: Cheap vs costly fuels. eg. Coal vs petrol.Explanation:Pollution: Clean fuel vs polluting fuel. Electricity, solar cooker, LPG are clean fuels, as they do not produce smoke while wood and coal fall in the second category.Choice of fuel in different conditions:A. In a forest: In the forest, we can not have electricity or transport gas cylinder. In such places, woods from the forest is easy to procure.B. In a remote mountain village or small island:Here also we can not get electricity or LPG CONNECTION. Since this is a mountainous region or island. Here we can easily get ENOUGH solar energy, so, we can use a solar cooker in such places.C. New Delhi:In a big city like Delhi, Mumbai electricity and LPG connection are available with good service. So we prefer them to cook our food. In Delhi, electricity is cheap so the use of electrical appliances like induction or heater incurs a lesser cost.D. Lived five centuries ago: A century has a life span of 100 years. Presently we live in the 21ST century. In the 16th century (1500-1599 AD) there was no fossil fuel, LPG or electricity. Peoples of 16th century cooked food in a clay stove using the energy of wood, coal and cow dung cake.How are factors different in each case:In each of the given situation, we first look for the availability of various fuel options. For example in forest wood is the only option to cook food. If more than one options are available, we use an easy to use and clean fuel. For example, in homes, we use LPG which is easy to use fuel and does not produce smoke. If all the available options are easy to use and clean we look for the cost factor, efficiency and other measures. |
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| 14. |
A spherical conductor of radius50 cm has a surfacecharge density of8.85 x 10°C/m². Theelectric field near thesurface in N/C is |
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Answer» re the Electric field near the surface is 10⁶ N/C.Given : A spherical CONDUCTOR of radius 50 cm has a surface charge density of 8.85 x 10¯⁶ C/m².To FIND : The electric field near the surface in N/C.solution : surface charge density, σ = 8.85 × 10¯⁶ C/m² so charge on the spherical conductor, Q = 4πr²σnow electric field DUE to sphere, E = kQ/r²= Q/(4πεr²)= (4πr²σ)/(4πεr²)= σ/ε= (8.85 × 10¯⁶)/(8.85 × 10¯¹²)= 10⁶ N/C THEREFORE the Electric field near the surface is 10⁶ N/C |
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| 15. |
An electron is moving on a circle of radius 0.5Å with a frequency of 5 × 10^16 Hz in anti-clockwise direction . The equivalent loop - current is? A)4.0mA (clockwise)B)4.0mA(anti-clockwise)C) 8.0mA(clockwise) D) 8.0mA(anti-clockwise) |
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Answer» (B) EXPLANATION:I don't KNOW the explanation |
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| 16. |
an object is placed infront of the concave mirror of focal length f.the magnification of the image becomes twice when the object distance is changed from 20cm to 10cm .Calculate the value of f. |
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Answer» ANSWERIt is given that,Focal lengthf=−20cmhi=3hoThe IMAGE is three TIMES the size of the object. So there are two cases.Case 1. For REAL imagem=−uv=−3v=3uUsing mirror’s formulaf1=v1+u1−201=3u1+u1u=3−80cmCase 2. For virtual imagem=u−v=+3v=−3uAgain using mirror’s formula−201=−3u1+u1u=3−40cmThe possible values of u are 3−80cmand3−40cm. |
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| 17. |
A car starts from rest and accelerates at 5 m/s through a distance of 22,5m. How fast is it they going?Hoe long did it take to cover 22,5 m ? |
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Answer» This may help youExplanation: Avg speed × total times = total distance20×25=500Since acceleration in the 1st interval is same as DECELERATION in the 3rd interval and deceleration is from the same VELOCITY achieved after acceleration, t 1=t 3s 1= 2 1 ×5t 1 s 2=(5t 1)t 2s 3=(5t 1)×t 3− 2 1×5×t 1 2 =(5t 1)×t 1− 2 1×5×t 1 2s 1+s 2+s 3=5t 1T 2+5t 1t 2⇒5t 1 2+5t 1 2=500−−−−12t 1+t 2=25−−−−2Solving (1 ) and (2) ⇒t 1=5,t 2=15sec |
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| 18. |
a particular is revolving in a circular path of radius 1:π m it distance after completing two and half rotation |
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Answer» ooooooioooooooooiiooiiiiiiiiiiiiiiipajsurhs8ooooo |
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| 20. |
An object is projected vertically up with 40 m/s from the ground. Its velocity after 6 sec of projection is (g = 10 2 / ms) |
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Answer» sec, A point of height at which object has no velocity means V=0 then time, V=U-gt , 40=0-10×t , then t is 4 sec( here - sign is indicated the direction of g ), it REACHES maximum height in 4 sec then after this it moved to ground with initial velocity is zero and after 2 SECOND it's velocity is ,V=U+gt , V=0+10×2 , V= 20m/sec is your ANSWER |
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| 21. |
Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius im surrounding the total charge is 100V-m. The flux over the concentric sphere of radius 2m will be: |
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| 22. |
If a body starts from rest and it acquires a velocity 20m/sec in 15sec, then acceleration of the body is |
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Answer» ecause BODY start from REST), V= 20 m/sec , t=15 sec , V=U+a×t , 20=0+a×15 , 20/15=a, a=4/3 m/sec^2 |
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| 23. |
Three small charges A (+2×10^-8), B(-5×10^-8) and P (+1×10^-8) lie along a line with distances AB= 6CM, BP= 4Cm and Ap=10 cm. (a) Calculate the force on the charge at P due to A and B.(b) At which point X on the line connecting A and B could there be no force on the charge P due to A and B if P were placed there?.Please no spaming |
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Answer» The Force on Charge P is 6.75 × 10⁻⁵ NThe Point X is X = 2.32 mGiven:Charge on A = 2 × 10⁻⁸ CCharge on B = - 5 × 10⁻⁸ CCharge on C = 1 × 10⁻⁸ CExplanation:As given the Distance of Separation are,⇒ AB = 6 cm = 0.06 m⇒ BP = 4 cm = 0.04 m⇒ AP = 10 cm = 0.1 MTHE Charges are kept as shown in figure:-Now, we can see that Force on P due to A will be repulsive as they are of same signs. and, Force on P due to B will be attractive as they are of opposite signs.As the Forces are ACTING at 180° angle the Resultant force will be the Subtraction of two forces,Now, lets find the Force F₁,From the FORMULA,⇒ F₁ = K (Q_{a} Q_{p}) / r²Substituting the values,⇒ F₁ = K × (2 × 10⁻⁸ × 1 × 10⁻⁸) / (0.1)²⇒ F₁ = K × (2 × 10⁻¹⁶) / (10⁻²)⇒ F₁ = K × 2 × 10⁻¹⁶ × 10²⇒ F₁ = 9 × 10⁹ × 2 × 10⁻¹⁶ × 10²⇒ F₁ = 9 × 10¹¹ × 2 × 10⁻¹⁶⇒ F₁ = 18 × 10⁽¹¹⁻¹⁶⁾⇒ F₁ = 18 × 10⁻⁵⇒ F₁ = 1.8 × 10⁻⁴⇒ F₁ = 1.8 × 10⁻⁴ NNow, Finding F₂ value,⇒ F₂ = K (Q_{b} Q_{p}) / r²Substituting the values,⇒ F₂ = K × (5 × 10⁻⁸ × 1 × 10⁻⁸) / (0.04)²⇒ F₂ = K × (5 × 10⁻¹⁶) / (4 × 10⁻²)⇒ F₂ = K × 5 × 10⁻¹⁶ × 10² / 4⇒ F₂ = 9 × 10⁹ × 5 × 10⁻¹⁶ × 10² / 4⇒ F₂ = 9 × 10¹¹ × 5 × 10⁻¹⁶ / 4⇒ F₂ = 45 × 10⁽¹¹⁻¹⁶⁾ / 4⇒ F₂ = 45 × 10⁻⁵ / 4⇒ F₂ = 11.25 × 10⁻⁵⇒ F₂ = 1.125 × 10⁻⁴⇒ F₂ = 1.125 × 10⁻⁴ NNow, let's Find the resultant force.⇒ F = F₁ - F₂(Here F Denotes Net Force)⇒ F = 1.8 × 10⁻⁴ - 1.125 × 10⁻⁴⇒ F = 10⁻⁴ (1.8 - 1.125)⇒ F = 10⁻⁴ × 0.675⇒ F = 6.75 × 10⁻⁵ ⇒ F = 6.75 × 10⁻⁵ N∴ The Force on Charge P is 6.75 × 10⁻⁵ N.Now, Point P is between A and B and we know the distance between A and B is 6 cm.Let the Point P be at X cm Distance from A and 6 - X cm Distance from BWe Know that Net Force is Zero. the force due to A is equal to force due to B.⇒ F₁ = F₂⇒ K (Q_{a} Q_{p}) / r₁² = K (Q_{b} Q_{p}) / r₂²⇒ (Q_{a} Q_{p}) / r₁² = (Q_{b} Q_{p}) / r₂²⇒ 2 × 10⁻⁸ × 1 × 10⁻⁸/(X)² = 5 × 10⁻⁸ × 1 × 10⁻⁸/(6-X)²⇒ 2 × 10⁻¹⁶ / X² = 5 × 10⁻¹⁶ / (6-X)²⇒ 2 / X² = 5 / (6 - X)²⇒ 2 × (6 - X)² = 5 × X²⇒ 2 × {36 + X² - 12 X} = 5 X²⇒ 2 X² - 24 X + 72 = 5 X²⇒ 3 X² + 24 X - 72 = 0Lets Find the Discriminant,⇒ D = b² - 4 a c⇒ D = (24)² - 4 × 3 × - 72⇒ D = 576 - 12 × - 72⇒ D = 576 + 864⇒ D = 1440.Now, Finding X Value by Quadratic Formula,⇒ X = ( - b ± √{D} ) / 2 a⇒ X = ( - 24 ± √{1440} ) / 2 × 3⇒ X = ( - 24 ± 12√{10} ) / 6⇒ X = 6 × (- 4 ± 2√{10} ) / 6⇒ X = - 4 ± 2√{10}Now,⇒ X = - 4 + 2√{10}(Here distance cannot be negative)⇒ X = - 4 + 6.32⇒ X = 2.32⇒ X = 2.32 cm∴ The Point X from A is X = 2.32 m. |
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| 24. |
An object is placed at 60 cm from a concave lens. The image is formed at a distance of 20 cm from the optical center. Find the focal length of the lens? Is the lens coverging or diverging? One who wants brainlist mark, answer this question. I will follow you..... |
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Answer» he lens is concave the object distance and the IMAGE distance, both will be negative.Given that:u = -60 cmv = -20 cm We know that:1/F = 1/v - 1/u=> 1/f = -1/20 + 1/60=> 1/f = (-3+1)/60=> 1/f = -2/60=> 1/f = -1/15Thus, f = -15 cm SINCE the lens if concave, it is a DIVERGING lensThanks! |
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| 25. |
An object is placed at 0.08 from a convex mirror of focal length 0.1 calculate the position and nature of the image. One who wants brainlist mark, answer this question. |
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Answer» focal LENGTH is positive for convex and NEGATIVE for concave.object and IMAGE if has same SIGN then the image is virtual.object and image if has different sign then the image is real |
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| 26. |
Hiiii every Good morning i want physics answer Question:-The magnification product by a mirror is +1. What does it mean? |
| Answer» ANSWER. The magnification produced by a PLANE mirror is +1 means that the image formed is VIRTUAL, ERECT and of the same size as that of object. | |
| 27. |
৪. একজন লােক দক্ষিণদিকে 2 m/s বেগে হাটছে। তার মনে হল ঠিক পূর্বদিক থেকে বায়ুপ্রবাহ হচ্ছে। লােকটি দ্বিগুণ বেগে চললে তার মনে হয় ঠিক দক্ষিণ-পূর্বদিক থেকে বায়ুপ্রবাহ হচ্ছে। বায়ুপ্রবাহের বেগ নির্ণয় করাে।[22 m/s , উত্তর-পূর্বদিক থেকে] |
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Answer» বায়ুপ্রবাহের নিয়ম ও বায়ুপ্রবাহের শ্রেণিবিভাগ :- বায়ু কয়েকটি নির্দিষ্ট নিয়মে প্রবাহিত হয় ।[১] চাপের সমতা রাখার জন্য বায়ুপ্রবাহ উচ্চচাপ অঞ্চল থেকে নিম্নচাপ অঞ্চলের দিকে ধাবিত হয় । বায়ুচাপের তারতম্য হল বায়ুপ্রবাহের প্রধান কারণ । বায়ুর চাপ আবার নির্ভর করে উষ্ণতর উপর । কোনো স্থানের বায়ু উত্তপ্ত হলে সেখানে বায়ুচাপ কমে যায় এবং আংশিক স্থান বায়ুশূন্য হয় । এই শূন্যস্থান পূরণের জন্য শীতল উচ্চচাপ যুক্ত স্থান থেকে বাতাস নিম্ন চাপের দিকে ধাবিত হয় । এইভাবে বায়ুপ্রবাহের সৃষ্টি হয় । [২] বায়ুচাপের পার্থক্যের ওপর বায়ুপ্রবাহের গতিবেগ নির্ভর করে । বায়ুচাপের পার্থক্য বেশি হলে বায়ুপ্রবাহের বেগ বৃদ্ধি পায়, অনুরূপভাবে বায়ুচাপ কমে গেলে বায়ুপ্রবাহের বেগও কমে যায় ।[৩] পৃথিবীর আবর্তন গতির জন্য বায়ুপ্রবাহ উত্তর গোলার্ধে ডানদিকে ও দক্ষিণ গোলার্ধে বামদিকে বেঁকে যায় । বায়ুপ্রবাহ কখনও উত্তর থেকে দক্ষিণে সোজাসুজি প্রবাহিত হয় না । এর কারণ[i] পৃথিবীর আবর্তনের বেগ নিরক্ষরেখার থেকে দুই মেরুর দিকে ক্রমশ কমে গেছে । [II] পৃথিবী পশ্চিম থেকে পূর্ব দিকে আবর্তন করছে ।[III] ওপরের বায়ুপ্রবাহের চেয়ে ভূপৃষ্ঠের সংলগ্ন বায়ুপ্রবাহ ভূপৃষ্ঠের সঙ্গে সংঘর্ষের ফলে কিছুটা পিছিয়ে পড়ে । বায়ু প্রবাহের এই গতি বিক্ষেপ [Deflextion] ফেরেলের সুত্র [Ferrel's LAW] নামে পরিচিত । [৪] বায়ুপ্রবাহের গতিপথে পাহাড়-পর্বত থাকলে বায়ুপ্রবাহ তার গতিপথ পরিবর্তন করে ।[৫] নিম্নচাপের হালকা বায়ু উপরের দিকে এবং উচ্চচাপের ভারী বায়ু নীচের দিকে প্রবাহিত হয় । ফেরেলের সূত্র [Ferrel's Law]:- অভিগত গোলাকৃতির পৃথিবী নিজের অক্ষের ওপর পশ্চিম থেকে পূর্বদিকে আবর্তন করছে, কিন্তু পৃথিবীর আবর্তন বেগ সর্বত্র সমান নয় । নিরক্ষরেখা থেকে দুই মেরুর দিকে আবর্তনের গতিবেগ ক্রমশ কম । আবার ভূপৃষ্ট সংলগ্ন বায়ু ভূপৃষ্ঠের সঙ্গে ঘর্ষণের ফলে ওপরের বায়ু থেকে কিছুটা পিছিয়ে পড়ে । এইসব কারণে পৃথিবীর ওপর কোনও গতিশীল পদার্থ, যেমন: বায়ুপ্রবাহ বা সমুদ্রস্রোতের গতির দিক সোজাসুজি উত্তর-দক্ষিণে না হয়ে উত্তর গোলার্ধে ডানদিকে ও দক্ষিণ গোলার্ধে বাঁদিকে বেঁকে যায় । বায়ুর গতি বিক্ষেপের এই নিয়মটি ফেরেলের সূত্র নামে পরিচিত ।ফেরেলের সূত্র অনুসারে:-১) পৃথিবী স্থির থাকলে উত্তর গোলার্ধে যে আয়নবায়ু সরাসরি উত্তর থেকে দক্ষিণে প্রবাহিত হত, পৃথিবী আবর্তন করে বলে তা উত্তর গোলার্ধে ডানদিকে বেঁকে উত্তর-পুর্ব আয়নবায়ু হিসাবে প্রবাহিত হয় । একই ভাবে,২) পৃথিবী স্থির থাকলে যে বায়ু দক্ষিণ গোলার্ধে দক্ষিণ থেকে উত্তর দিকে সোজাসুজি প্রবাহিত হত, তা পৃথিবীর আবর্তনের জন্য খানিকটা বাঁদিকে বেঁকে দক্ষিণ-পূর্ব দিক থেকে প্রবাহিত হয়, যা দক্ষিণ-পূর্ব আয়নবায়ু নামে পরিচিত । দক্ষিণ-পূর্ব আয়নবায়ু নিরক্ষরেখা অতিক্রম করার পর উত্তর গোলার্ধে ডান দিকে বেঁকে দক্ষিণ-পশ্চিম মৌসুমি বায়ু হিসাবে প্রবাহিত হয় ।বায়ুপ্রবাহের শ্রেণিবিভাগ :- বায়ুপ্রবাহ প্রধানত ৪ রকমের হয়, যথা:- [১] নিয়ত বায়ুপ্রবাহ, [২] সাময়িক বায়ুপ্রবাহ, [৩] আকস্মিক বায়ুপ্রবাহ এবং [৪] স্থানীয় বায়ুপ্রবাহ ।Explanation: |
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| 28. |
Is it necessary to change the units into SI unit while using equation of motion? |
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Answer» The equations of MOTION, also known as SUVAT equations, are used when ACCELERATION, a, is constant. ... The quantities s, u, v and a are all vector quantities so therefore their sign represents the DIRECTION of motion. You NEED to ensure that the measurements are in base SI units before substituting them into the formulae.PLEASE CHOSE AS BRAINLIESTExplanation: |
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| 29. |
Solve the problem gsgdjffiif |
| Answer» PLEASE PROVIDE APPROPRIATE QUESTION | |
| 32. |
5 metre how many centimeters |
| Answer» | |
| 34. |
Angle of an inclined plane is tethA with the horizontal. From the foot of this plane a body is projected up the plane at an angle a with the plane. For maximum range a |
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Answer» afjeysyotkdykpyld, I |
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| 35. |
A car cover 3 km in 1 hr and 12 km in 2 hr in a straight line path it's average velocity is.. |
| Answer» AVG velocity= total displacement/ total TIME taken. = 3+12km/1+2hrs = 15km/3hrs= 5km/hr ih the avg velocity.pls mark it as the BRAINLIEST ANSWER | |
| 36. |
In a cuboid block ABCD is an area making with horizontal length, net stress on the surface ABCD (side AB and CD are parallel to upper edges and equal tolength / of square cross-section of block) |
| Answer» BC and DA is a SECTION of BLOCKS | |
| 37. |
Correct the following sentences active to passiveHe were told to keep quiet |
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Answer» he was TOLD to KEEP quietExplanation:MARK as BRAINLIEST PLEASE |
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| 38. |
An athlete complete one round of circular park in 40s. The radius of circular park is 77m. He moves on it for 100s. Find distance and displacement covered by him. |
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Answer» 1 round COVERED in 40 secondsso in 100 seconds he will move 5/2 round i.e. 2.5 ROUNDS of trackso as the RADIUS is = 77Distance = (2pi r = 1 round ) so 2.5 round = 2pi r * 5/2 = 5pi rDisplacement = 2r as at LAST he has only COMPLETED half round so he will be on the other side od diameter of circle so = 2r = 154mExplanation: |
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| 40. |
an object cover 30 m due east and the cover40 m due north then the ratio of its distance displacement is |
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Answer» 7:5 is the ANSWER USING PYTHAGORAS THEOREM |
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| 41. |
काँच की एकप्लेट पर प्रकाशकीएक किरण 30° केआपतन कोण के साथ गिरती है, यदि हवा के संबंध में काँचका अपवर्तनांक 1.5 है, तो अपवर्तन कोण है:(a) sin-(0.33) (b) sin-'(0.67)(c) sin-'(0.45) (d) sin-(0.03) |
| Answer» AEXPLANATION:because when LODA become LONGER then answer BECOMES A | |
| 42. |
Why in the above question distance get doubled |
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Answer» (A):-F/4Explanation:PLEASE THANKS me and MARK me as brainlist.Please it is my HUMBLE REQUEST. |
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| 43. |
end of 20th century there would be a single power dominating the world in this context what would be the role of |
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Answer» This article provides a single integrated narrative of the CHANGING CONTEXT of WORLD politics, from the outbreak of World War I to the 1990s.please FOLLOW me |
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| 44. |
Find the pressure at the depth of 0.6m in water of density 1000kg/m cube. (take g=10m/s square) plz say fast.. |
| Answer» DING to GIVEN QUESTION : | |
| 45. |
What is catenation and micille |
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Answer» Explanation: CATENATION is the BONDING of atoms of the same element into a series, CALLED a CHAIN. Micelle is an aggregate of SURFACTANT molecules dispersed in a liquid colloid. |
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| 46. |
Р at equilibrium and P is total9For the reaction equilibrium, 2NOBr(9) 2NO(g) + Br (9) If Ps.K,pressure, the ratio of is equal toР9(2)8111273New Delhi-110005 Ph 011-47623456 |
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Answer» asduuxbbvfnjy121234567890qwertyuiopasdfghjklzxcvbnm1+×÷=%_€£¥₩[EMAIL PROTECTED]#$/^&*()☆▪¤《》¡¿♧◇♡♤■□●○•°][}{><|\~`☆ |
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| 47. |
After rounding off the number 9367 to 2 significant digits the value becomes how much |
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Answer» 9400Explanation:PLEASE THANKS me and MARK me as brainlist.Please it is my HUMBLE REQUEST. |
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| 48. |
How a turbo chared works ? |
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Answer» The turbocharger on a car APPLIES a very similar principle to a PISTON engine. It uses the exhaust gas to drive a turbine. This SPINS an air compressor that pushes extra air (and OXYGEN) into the cylinders, allowing them to burn more fuel each second.I HOPE it helps youplz Follow me and Mark me as BRAINLIEST |
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| 49. |
When an air bubble comes upwards from bottom of a silent lake slowly its size (A) increases(B) decreases(C)remains same(D)none |
| Answer» | |
| 50. |
A car moving with a velocity of 10 m/s accelerates for 5 s. it covers a distance of 100 m during this time interval. Find a) the a. b) the v at the end of 5 s. |
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Answer» a =6m/s,v=10 m/sExplanation:here we are given U= 10m/sT=5sS=100mA=?APPLY SECOND equation of motionyou will get 6.now apply first equation of motionand you will get 40. I hope this answer was helpful if yes so PLEASE mark the answer as brainliest follow me and click that thanks BUTTON ✌️✌️✌️✌️✌️✌️ |
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