Explore topic-wise InterviewSolutions in Current Affairs.

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.

Deduse the dimensional formule of the following physical quantities .1.specific heat 2. mechanical equivalent of heat 3.cofficient of viscosity 4.time period5.Relative density 6.Angular speed 7.a pure number ​

Answer»

ANSWER:

That's all.All the BEST.
Discussion
2.

3) Match the following.[1] Copper + Hydrogen-- Copper Oxide a) Reduction2) Copper Oxide + hydrogen -- copper +water b) oxidation 3 lead Oride + Carbon--- lead + carbon c)Redox Reactiona)1-b.2-a.3-cb)1-a. 2-b.3-cc)1-c.2_b.3-ad)1-c.2-a.3-b ​

Answer»

Answer:

a)

EXPLANATION:

Reduction: Is KNOWN as losing oxygen.

Oxidation: Is known as losing Hydrogen.

Redox reaction : one reactant under GOES oxidation and ANOTHER under goes Reduction reaction.
Discussion
3.

Find the cosine of the angle between the given vectors: P = 3î + 12ſ – 4k and 7 = 2î + 2ỉ + k.Ans. cos O = 2/3solution of this ??​

Answer»

ANSWER:

my ANS is COS ∅ = 2/13.......
Discussion
4.

Prove that A = (4î – 6ị + 8k) and B = (2î + 4ſ + 2k) are mutually perpendicular.Calculate the work done​

Answer»

ANSWER:

HI

Explanation:

PLEASE follow me

L am new in there
Discussion
5.

A body of mass 10 kg is lying on a rough plane inclined at an angle of 30° to the horizontal and the coefficient of friction is 0.5. the minimum force required to pull the body up the plane isA. 914 N B. 91.4 N C. 9.14 N D. 0.914 N​

Answer»

Answer:

91.4 N

Explanation:

Force needed = Friction (UsN) + MG sin theta

= 25 root 3 + 50 N

approx 92 if u take  a due to g as 10m/s^2
Discussion
6.

An electron of mass m and charge e initially at rest gets accelerated by a constant electric field E.The rate of change of de broglie wavelength of this electron is​

Answer»

Answer:

around 0.01 NANOMETERS

Since electrons have a rest mass, unlike PHOTONS, they have a de BROGLIE wavelength which is really short, around 0.01 nanometers for easily achievable speeds. This means that a microscope using electron "matter WAVES" instead of photon light waves can SEE much smaller things.
Discussion
7.

Show that if the dot product of two finite vectors is zero then they are mutually perpendicular toeach other.​

Answer»

ANSWER:

hiiiii

Explanation:

L am NEW in there

PLEASE FOLLOW me
Discussion
8.

4. Manasvi moves with speed of 6 km/h for 2 h and with speed of 4 km/h for next 3 h. find the averagespeed of Manasvi and total displacement moved.Art. 2C11​

Answer»

Answer:

PLEASE maaf me as a brainliest

Explanation:

SPEED is EQUAL to distance into time therefore

6km =distance ×2h

distance =6/2=3m

now

4km=distance ×3h

distance =4/3=1.3

average speed =6-4 =2ms-²

displacement=2.7m

Discussion
9.

Lale 01 UNDIVIULUNIf the acceleration of a moving object is constant then the motion is saidto be[(a) Motion with Constant Speedted experimentsduct an expermiment to find acceleration and velocity of an object movingined plane and write a report.ed projectsalate the avarage speeds of students of your class we who have participaneters and 200 meters running race. Write a report.(b) Motion with Uniform Acceleration(c) Motion with Uniform Velocity (d) Motion with Non Uniform acceleratio​

Answer»

ANSWER:

If the acceleration of a moving object is CONSTANT then the MOTION is SAID (b) Motion with Uniform Acceleration.
Discussion
10.

The speed of the light in diamond is 1, 24, 000 km/s. Find the refractive index ofdiamond if the speed of light in air is 3, 00,000 km/s.

Answer»

\bigstar Given

The speed of the light in a diamond is 1,24,000 km/s

The speed of light in AIR is 3,00,000 km/s

\bigstar To Find

The refractive INDEX of diamond

\bigstar Solution

We KNOW that

\red{\sf \longrightarrow n=\dfrac{c}{v}}

Here

  • n = refractive index of MEDIUM
  • c = speed of light in air
  • v = velocity of light in medium

According to the question

Hence

The medium is diamond

And

We are asked to find the the refractive index of diamond

Therefore

We must find "n"

Given that

The speed of the light in a diamond is 1,24,000 km/s

The speed of light in air is 3,00,000 km/s

Hence

  • v = 1,24,000 km/s
  • c = 3,00,000 km/s

Substituting the values

We get

\green{\sf \longrightarrow n=\dfrac{3,00,000}{1,24,000}}

On further simplification

We get

\orange{\sf \longrightarrow n=\dfrac{300}{124}}

Hence

\pink{\sf \longrightarrow n=2.4}

Therefore

The refractive index of diamond = 2.4
Discussion
11.

ਪ੍ਰਸ਼ਨ 8. ਕਿਹੜੀ ਸੂਚਨਾ ਤੁਕ ਪੜ੍ਹਨ ਤੋਂਬਾਅਦ ਛੋਟਾ ਅਰਾਮ ਨਾਲ ਬੈਠ ਗਿਆ?2 points*®) Railway is a public property helpto maintain itOO21) A proper queue helps earlydisposal) The ticket office remains openday and night0H) Retiring rooms are booked at​

Answer»

ANSWER:

oroewpktnwofnsjsodejfjejrjs
Discussion
12.

2.A rod of length 1m and mass 0.5 kg hinged at one end, is initially hanging vertical. The other end is nowraised slowly until it makes an angle 60° with the vertical. The required work is: (use g =9,8 m/(A) 1.522 J(B) 1.225 J(C) 2.125 J(D) 3.125 K​

Answer»

hope this will be HELPFUL to you

please FOLLOW me only if U like
Discussion
13.

A solid weight 32gf in air and 28.8gf in water. Find the volume and RD of the solid. Find how much it will weigh in a liquid of density 0.9 g/cm³.please give the answer with process in a paper irrelevant answers will be reported please help friends​

Answer»

ANSWER:

Hii..!

{\huge{\tt{\boxed{\fcolorbox{red}{black}{Solution:-}}}}}

_____________________________________

Given,

Weight of solid in air, W, 32 gf

Weight of solid when COMPLETELY IMMERSED in water W2 = 28.8 gf

(i) VOLUME of solid = Mass / density of solid.

= \frac{32}{10} = 3.2 {m}^{2} \\

(II) R.D. of solid =

\frac{w1}{w1 - w2} \times r.d \: \: of \: \: water. \\ \\ \\ r.d \: \: of \: \: solid \: = \: \frac{32}{32 - 28.8} \times 1 \\ \\ \\

So, R.D of water = 10.

Hence Solved!

_____________________________________

Hope it helps you dear!
Discussion
14.

Q.7 A wire of length L and 3 identical cells of negligibleinternal resistance are connected in series. Due to thecurrent, the temperature of the wire is raised by AT ina time t. A number N of similar cells is now connectedin series with a wire of the same material and cross-section but of length 2L. The temperature of the wire israised by the same amount AT in the time. The value of N is (a) 4. (b)6. (c)8. (d)9. ​

Answer»

ANSWER:

jejskwkekwk4jekrjdjfj4jsjajdjfhrsurjdisj3neuwieirudjejddjejdieid7euduejdjdjeudjeidjejdjeje

endnf

Explanation:

jsjejdjeejueueruejeu
Discussion
15.

What is the vertical component of a displacement vectorS = 26m makes angle 60 degreesang with X-axis?​

Answer»

26sin60

26×√3/2

13√2

hope it HELPS you

Discussion
16.

HOTEGER28.A cloth strip dipped in onion juice is used for testing a liquid %. Theliquid 'X' changes its odour. Which type of an indicator is onion juice?The liquid 'X' turns blue litmus red. List the observations the liquid Xwill show on reacting with the following:Zinc granules(b)Solid sodium carbonateWrite the chemical equations for the reactions involved.​

Answer»

Answer:

ok .

just WRITE the ELEMENT NAME and MULTIPLY by 2 ..
Discussion
17.

The period of revolution of the moon around the earth is 27.3 days. Rememberthat this is the period with respect to the fixed stars (the period of revolutionwith respect to the moving earth is about 29.5 days; it is this period that isused to fix the duration of a month in some calendars). The radius of moon’sorbit is 3.84 × 10^8m (60 times the earth’s radius). Calculate the centripetalacceleration of the moon and show that it is very close to the value given by9.8 ms–2 divided by 3600, to take account of the variation of the gravity as 1/r2

Answer»

ANSWER:

now what U need in this QUESTION of others bro
Discussion
18.

The displacement of a particle moving along x-axis is given by X = 18t + 5t.calculate:(i) the instantaneous velocity at t = 2s.(ii) average velocity between t = 2 s and t = 3s.(iii) instantaneous acceleration​

Answer»

GIVEN

The displacement of a particle moving along X-axis is given by

  • x = 18t + 5t²

To Find

i. Instantaneous velocity at t = 2 s

ii. Average velocity b/w t = 2 s and t = 3 s

iii. Instantaneous ACCELERATION

Solution

i.

\rm v=\dfrac{dx}{dt}\\\\\implies \rm v=\dfrac{d}{dt}(18t+5t^2)\\\\\implies \rm v=18+10t

Instantaneous Velocity at  t = 2 s is given by ,

\implies \rm v=18+10(2)\\\\\implies \bf \pink{v=38\ m/s}\ \; \bigstar

ii.

Average velocity is given by displacement per change in time

\implies \rm A_{v}=\dfrac{x_2-x_1}{t_2-t_1}

\implies \rm A_v=\dfrac{\{18(3)+5(3)^2\}-\{18(2)+5(2)^2\}}{3-2}\\\\

\implies \rm A_v=\dfrac{\{54+45\}-\{36+20\}}{1}\\\\\implies \rm A_v=99-56\\\\\implies \bf \green{A_v=43\ m/s}\ \; \bigstar

iii.

Instantaneous acceleration is given by ,

\implies \rm a=\dfrac{dv}{dt}\\\\\implies \rm a=\dfrac{d}{dt}(18+10t)\\\\\implies \bf \blue{a=10\ m/s^2}\ \; \bigstar
Discussion
19.

A mass m sliding horizontally is subject to a viscous drag force. For an init ial velocity v0 (at x = t = 0) anda retarding force F = - bx find t he velocity as a function of distance, v(x) , and show t hat the mass moves afinite distance before coming to rest

Answer»

The velocity v(x) as function of x is:

\boxed{v(x)^2=v_0^2-\frac{bx^2}{m}}

The finite distance moved by the mass is:

\boxed{x=v_0\sqrt{\frac{m}{b}}}

Explanation:

Given

Retarding force

F=-bx

Initial velocity

u=v_0

The mass is m

Acceleration due to retarding force will be

a=\frac{F}{m}

\implies a=\frac{-bx}{m}

We know that

a=\frac{dv}{dt}

or a=\frac{dv}{dx}.\frac{dx}{dt}

or, a=v\frac{dv}{dx}

Thus,

v\frac{dv}{dx}=-\frac{bx}{m}

\implies mvdv=-bxdx

\implies m\int_{v_0}^{v(x)} vdv=-b\int_0^x xdx

\implies m\frac{v^2}{2}\Bigr|_{v_0}^{v(x)}=-b\frac{x^2}{2}\Bigr|_0^x

\implies m(\frac{v(x)^2}{2}-\frac{v_0^2}{2}=-b\frac{x^2}{2}

\implies \boxed{v(x)^2=v_0^2-\frac{bx^2}{m}}

This is the expression of velocity v(x) as a function of distance x

Now if v(x)=0

Then

0=v_0^2-\frac{bx^2}{m}

\implies x^2=\frac{mv_0^2}{b}

\implies x=\sqrt{\frac{mv_0^2}{b}}

\implies \boxed{x=v_0\sqrt{\frac{m}{b}}}

Which is the finite distance moved by mass before coming to rest.

Hope this answer is helpful.

Know More:

Q: When a force act on a BODY of mass its position x VARIES with TIME t as x=at^4+bt+c where a,b,c are CONSTANTS workdone is:

Click Here: brainly.in/question/16376510
Discussion
20.

Given ā = i + 2; and b = 2i + jwhat are the magnitudes of the two vectors? Are these two vectors equal​

Answer»

= \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos( \theta) }

\sqrt{ {3}^{2} + {3}^{2} + 18 }

= \sqrt{36}

= 6
Discussion
21.

Calculate work done on charge 1Cin an electrostatic field 2 Newton/c alongx axis when charge is moved the through adistance2.5 m along+x-axis ​

Answer»

we have to CALCULATE the work done on charge 1 C in an electrostatic field 2 N/C along x - axis when charge is moved through a distance 2.5 m along +x axis.

solution : we KNOW force, F = qE

where q is charge and E is electrostatic field.

here q = 1 C and E = 2 N/C along x - axis

so the force along x - axis , F = 1 C × 2N /C = 2 N

displacement along x - axis , s = 2.5 m

work done = force × displacement along force

= 2 N × 2.5 m

= 5 Nm = 5J

Therefore the workdone on the charge 1 C in an electrostatic field 2 N/C along x - axis is 5 J

Discussion
22.

1 )You are Arun/Anita Basu living at 1785, Sector 29, Chandigarh. You camne serons the followingPostal Classes-Diploma in Waste Water Management duration one year--practiunla arrangedminimum qualification SSCE first division-prospectus free-Contact Director, Eco-centre, P.B. 1927advertisement:Bengaluru-560019Write a letter to the Director asking for detailed information on the course, fee structure,assignments, personal contact programmes and job opportunities. Request for a prospectusand enclose a self-addressed envelope.(Word limit: 120-150) pls answer it fast..........and good ni8 to everyone​

Answer»

ANSWER:

560019

I hope answer is HELPFUL to us

PLEASE follow me and GIVE THANKS
Discussion
23.

Two resistors of resistance 3ohm and 6 ohm are connected to a battery of 6 V , so as to have minimum resistance. Show how will you connect them(diagram) and also find the current in this case.

Answer»

Given

Two resistors of resistance 3 OHM and 6 Ohm are connected to a battery of 6 V

To Find

How to connect them for getting MINIMUM resistance

Knowledge Required

When resistors are connected in series ,

\bf \bigstar\ \; \pink{R_{eq}=R_1+R_2+R_3+...}

When resistors are connected in parallel ,

\bf \bigstar\ \; \green{\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...}

Ohms Law :

\bf \bigstar\ \; \blue{V=IR}

where ,

V denotes potential DIFFERENCE

I denotes current

R denotes resistance

Solution

When 3 and 6 Ω are connected in series ,

\rm R_{eq}=3+6\\\\\implies \bf R_{eq}=9\ \Omega

When these are connected in parallel ,

\rm \dfrac{1}{R_{eq}}=\dfrac{1}{3}+\dfrac{1}{6}\\\\\implies \rm \dfrac{1}{R_{eq}}=\dfrac{3}{6}\\\\\implies \rm R_{eq}=\dfrac{6}{3}\\\\\implies \bf R_{eq}=2\ \Omega

So , We will get minimum resistance , when connected in parallel .

__________________________

  • Eq. Resistance = 2 Ω
  • Potential Difference = 6 V

Apply Ohms Law ,

\bf \red{\bigstar\ \; V=IR_{eq}}\\\\\rm \implies 6=I(2)\\\\\implies \rm I=\dfrac{6}{2}\\\\\implies \bf \orange{I=3\ A\ \; \bigstar}

So , Current = 3 A
Discussion
24.

Ii)a) 2H,BO3(aq)Given the following equations and A Hºvalues at 25° C,> B,0316) +3H2OAH° = +14.4 kJb) HBO3(aq)HBO+H2OAH° = -0.02 kJ176) 2B,036) + H2OAH° = 17.3 kJCalculate A Hº for the following reactionProgramme -GENIUS Learning!(1)c) H.B.076)at 298 K4HBO2(aq)H,B,0,0) + H2OAH = ?​

Answer»

ANSWER:

I can't UNDERSTAND your QUESTION
Discussion
25.

Which of the following device converts electric energy to electric current?(a) An electric cell(b) Electric wires(c) An electric bulb (d) A switch​

Answer»

ANSWER:

A) An ELECTRIC CELL.

HOPE the answer is correct✅
Discussion
26.

A trainacelerated from 10 km/her to 40 km/hin a 2 minutes. how much distance would itan this period​

Answer»

u = 10 \: km {}^{ - 1} = 2.7 \: ms {}^{ - 1}

v = 40 \: km {}^{ - 1} = 11.1 \: \: ms {}^{ - 1}

t = 2 \: minutes = 120 \: sec

a = ?

a = \frac{v - u}{t}

a = \frac{11.1 - 2.7}{120}

\huge\pink{\boxed{\purple{a = 0.07 \: ms {}^{ - 2}}}}

For Calculating Distance,

\huge\pink{s = ut + \frac{1}{2} at {}^{2} }

\large{s = 2.7 \times 120 + \frac{1}{2} \times 0.07 \times 120 \times 120}

\large{s = 324 + 504}

\huge\green{\boxed{\purple{s = 828 \: meter}}}

\huge\pink{\underline{\underline{\mathbb{\orange{BrAInLIesT}}}}}
Discussion
27.

1 mole propore (GH) on arbuston TPKilo Toules of heat eroagehehralod cohor - pepontonombuston at S.TP.​

Answer»

ANSWER:

THANKS for FREE POINTS
Discussion
28.

Round pff the following number to three significant figuresa.3.264b.0.9462c.1. 667d.1.285e.45.875​

Answer»

ANSWER:

b. 0.9462

Explanation:

ok BYE GOOD NIGHT
Discussion
29.

4. Match the following Lulus.Column 'A(a) Datum(b) Isobath(c) Topo sheet 45D/7(d) Telephone line(e) Hamlet() Canal(3) MetropolisColumn 'B'3. (i) Lines joining areas of equal depth(ii) Rural1. (ii) Mean sea level(iv) Sirohi4 (v) Communication(vi) Urban(vii) Irrigation​

Answer»

ANSWER:

I can't UNDERSTAND this SORRY
Discussion
30.

1. Calculate the weight of a 50kg mass on Earth (note g = 9.8N/kg)2. Calculate the weight of a 200kg mass on Earth3. On Jupiter the gravity measures 23N/kg. Calculate the mass of a rock which has a weight of 4600N.4. Calculate the weight that the same rock would have on Earth.5. A 20N force acts at an angle of 25 degrees to the horizontal. Resolve the force to find it's horizontal and vertical components

Answer»

Answer:

1. Weight is given by the FORMULA,

W=mg.

On EARTH, g=9.8 m/s2 . So, the object's weight on Earth will be:

W=50 kg⋅9.8 m/s2.

=490 N.

2. Calculate this ONE just as I did above

PLEASE, MARK ME AS BRAINLIEST!

Discussion
31.

B) A force of 2N is applied on a body of mass 10kg for 5s. Calculatechange in momentum, change in velocity and accelerationproduced in the body.​

Answer»

ANSWER:

F=M×A

HOW 2N = 10KG ×acceleration
Discussion
32.

प्रकृति का मूल बल नही हैगरुत्वाकर्षणविद्युतचुम्बकीयनाभिकीयअभिकेंद्रChoose only one please help​

Answer»

ANSWER:

maybe....... obtion (C)

EXPLANATION:

i can't understand hindi

Discussion
33.

1.How many significant figures are there in the following quantities ?a.10.163b.1. 67×10-17 c.0.270d.1.496e.15000f.2. 4300g.0.001040​

Answer»

ANSWER:

a.5

b.3

c.4

d.4

e.2

f.3

g.6
Discussion
34.

प्रकृति का मूल बल नही है —गरुत्वाकर्षणविद्युतचुम्बकीयनाभिकीयअभिकेंद्र​

Answer»

ANSWER:

nabhikiya............ OK
Discussion
35.

At 300K what is the rms speed of Helium atom?[ mass of He atom is 4u, 1u = 1.66 x 10-27 kg, kB = 1.38 x 10-23J/K]​

Answer»

Answer:-

\red{\bigstar}\large\leadsto\boxed{\sf \green{1368 \: m/s}}

• Given:-

Temperature [T] = 300 K

Mass of He atom = 4U

Boltzmann constant [KB] = 1.38 × 10-²³ J/K

• Solution:-

\pink{\bigstar}\boxed{\sf \red{v_{<klux>RMS</klux>} = \sqrt{\dfrac{3k_{B}T}{m}} = \sqrt{\dfrac{3RT}{M_{m}}}}}

where,

\sf{\dfrac{M_{m}}{m} = \dfrac{R}{kB}}

here,

kB = Boltzmann constant

T = Temperature

M = MOLAR mass

m = mass of gas in kg

R = Universal Gas constant

Hence,

\pink{\bigstar}\boxed{\sf \red{v_{RMS} = \sqrt{\dfrac{3RT}{M_{m}}}}}

TAKING:-

R = 8.314472 J/mol. K

T = 300 K

Mm = 4.0026 × 10-³ kg/mol.

→ \sf \sqrt{\dfrac{3 (8.314472 J/mol.K) 300 K}{4.0026 \times 10^{-3} kg/mol}}

→ \sf \sqrt{\dfrac{7483.0248}{4.0026 \times 10^{-3} kg/mol}}

→ \sf{\sqrt{1869.540998 \times 10^{3} kg/mol}}

→ \sf{\sqrt{1869540.998}}

→ \sf{1367.311595}

Therefore, the rms speed of He is 1367.311595 ≈ 1368 m/s.
Discussion
36.

Arespect toPasengaI 2 codechto say" withtravellingin а abut the passengertravelling inanothes but bus inthesame speed issame Linctionthewithat​

Answer»

don't KNOW bhai

Explanation:

PLEASE FOLLOW me
Discussion
37.

Take a ball and a piece of paper (rectangularshape). Drop them simultaneously from the first/second floor of the building. Observe whether both reach the groundsimultaneously. Now crush the same paper into spherical shape and again drop the paper and ball simultaneously from the first/second floor of the building. Again observe whether both reach the ground simultaneously. Repeat the same for four times. Prepare the report with discussion and conclusion of this activity.​

Answer»

ANSWER:

HEY JACK had OOO just it's y'all
Discussion
38.

I)Heat of fusion of ice at 0°C and 1atmosphere is 6.01 kJ mol" and heat ofevaporation of water at 100 °C is 40.7 kJmol-1. Calculate the enthalpy change forthe conversion of 1 mole of ice at 0°C intovapour at 100°C. Heat capacity of wateris 4.184 JK-1 mol1.​

Answer»

Answer:

heat RELEASED to cool 500g water from 20

∘

C to 0

∘

C.

q=msΔT

=500×4.18×20=41800J

=41.8kJ

Number of MOLES of water(ICE) that will melt to absorb 41.8kJ=

6.02

41.8

=≈7.
Discussion
39.

Derivation of first law and second law of motion Derivation of first law of motion

Answer»

Answer:

Newton's first law states that a body stays at rest if it is at rest and moves with a CONSTANT velocity unit if a net force is applied on it.

or

An object at rest will remain at rest unless acted upon by an external and unbalanced force. An object in motion will remain in motion unless acted upon by an external and unbalanced force.The first law gives the qualitative explanation of force.

it doesn't have that type any numerical application

Newton's second law states that the net force applied on the body is equal to the RATE of change in its momentum.

F = ma

a(acceleration) = v-u/t

or F = m(v-u) / t

or Ft = mv - mu

That is, when F = 0, v = u for whatever time, t is taken. This means that the object will continue moving with uniform velocity, u throughout the time, t. If u is ZERO than v will also be zero, i.e., object will remain at rest.

derivation of three equations of motion :

Here is your answer........!!

(1) First equation of Motion:

V = u + at

soln.

Consider a body of mass “m” having initial velocity “u”.Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”.

Now we know that:

Acceleration = change in velocity/Time taken

=> Acceleration = Final velocity-Initial velocity / time taken

=> a = v-u /t

=>at = v-u

or v = u + at

This is the first equation of motion.

—————————————-

(2) Second equation of motion:

s = ut + 1/2 at^2

sol.

Let the distance travelled by the body be “s”.

We know that

Distance = Average velocity X Time

Also, Average velocity = (u+v)/2

.: Distance (t) = (u+v)/2 X t …….eq.(1)

Again we know that:

v = u + at

substituting this value of “v” in eq.(2), we get

s = (u+u+at)/2 x t

=>s = (2u+at)/2 X t

=>s = (2ut+at^2)/2

=>s = 2ut/2 + at^2/2

or s = ut +1/2 at^2

This is the 2nd equation of motion.

……………………………………………………………

(3) Third equation of Motion

v^2 = u^2 +2as

sol.

We know that

V = u + at

=> v-u = at

or t = (v-u)/a ………..eq.(3)

Also we know that

Distance = average velocity X Time

.: s = [(v+u)/2] X [(v-u)/a]

=> s = (v^2 – u^2)/2a

=>2as = v^2 – u^2

or v^2 = u^2 + 2as

This is the third equation of motion.

Plz mark as brainlist.........!!

Explanation:
Discussion
40.

Two balls A and B of masses m and 3m are in motionwith velocities 3v and v respectively. Their inertia arein the ratio​ ( pls explain with steps )

Answer»

ANSWER:

SENSE INERTIA is a MEASURE of mass

So RATIO is 1:3 or 3:1
Discussion
41.

Toandoutesspace.correctto say"withrespecta bus, the passengerIs itthePassengerbut bus inspeed isthesametravelling intravelling in anothessameLirection withatrest"?​

Answer»

ANSWER:

U subdjebubUwbjqnIninobqzubxneixbeniebxxineienxrnxinrkxnrj
Discussion
42.

In cgs system F=100 dyne in other system of kg,metre , minute find the magnitude of forcetell correct answer i will mark as brainliest

Answer»

Your SOLUTION is in the PICTURE
Discussion
43.

Which of the following is vector quantity?a speedb timec all of these d velocity

Answer»

ANSWER:

time

Explanation:

SPEED is a SCALER QUANTITY

Discussion
44.

When white light is passed through a prism it causes dispersion. why does this happen??answer fast.....don't spam.....and no need of giving irrelevant answers.....​

Answer»

Dispersion of light: The splitting of WHITE light into seven colours on passing through a transparent medium is called dispersion of light. The dispersion of white light happens because the angle of REFRACTION of lights of DIFFERENT colours is different while passing through the transparent medium.

If you are SATISFIED so PLEASE mark as brainlist

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Discussion
45.

Mass is 210g and density is 7.981 g/cm^3 what will be the volume with regard to significant figurestell me correct answer you will be marked as brainliest

Answer»

ANSWER:

CHOOSE correct answer:

1.Bhirrana/Kalibangan has emerged as the oldest indus site.

2.The drains/streets cut each other at right angles.

3.Most of the seals were MADE of terracotta/bitumen.

4.Many/No TEMPLES have been found in the indus cities.

5.The indus people traded with Mesopotamina/Greece.

6.The indus people used Roman/pictographic script.
Discussion
46.

Which of the following electromagnetic waves has (a) minimumwavelength, and (b) minimum frequency ? Write one use of each of thesetwo waves.Infrared waves, Microwaves, y-rays and X-rays​

Answer»

Answer:

a) minimum WAVELENGTH

it HELPS in listening ULTRASOUND waves

a) infrared waves - it is USEFUL in COMMUNICATION.

b) microwaves - radio astronomy ,medical treatment

these are the two waves uses.

Discussion
47.

A ball is thrown at 20.0 m s−1 at an angle 20.0° above the horizontal. Calculate the size of the ball’s instantaneous velocity at its highest point.(You may ignore air resistance. Give your answer in m s−1 to an appropriate number of significant figures.)

Answer»

Given:

A ball is thrown at 20.0 m s−1 at an ANGLE 20.0° above the horizontal.

To find:

Instantaneous VELOCITY of the ball at the highest point.

Calculation:

This QUESTION is based on ground to ground Projectile.

We know that at the highest point of a projectile the instantaneous velocity of an object is equal to the x-axis component of the initial velocity.

\rm{ \therefore \: (v)_{highest \: point} = v_{x}}

\rm{ = > \: (v)_{highest \: point} = v \times \cos( \theta) }

\rm{ = > \: (v)_{highest \: point} = 20\times \cos( {20}^{ \circ} ) }

\rm{ = > \: (v)_{highest \: point} = 20\times 0.93 }

\rm{ = > \: (v)_{highest \: point} = 18.79 \: m {s}^{ - 1} }

So, final ANSWER is:

\boxed{ \red{ \rm{ \: (v)_{highest \: point} = 18.79 \: m {s}^{ - 1} }}}
Discussion
48.

08 संख्याएँ 2.05+2.005+2.0005 का योग होगा।(2 Points)2.05552.062.052.1​

Answer»

ANSWER:

6.0555

Explanation:

OPTIONS glat h OK fine

kutto mil gya na answer CHLI niklo dawfa HO jaao

Discussion
49.

A 2 kg ball is moving at a speed of 4 m/s toward a ball of the same mass at rest. They both stick together after the collision. What is the velocity of the two ball system and what type of collision is it?2 m/s, elastic collision5 m/s, inelastic collision2 m/s, inelastic collision5 m/s, elastic collision​

Answer»

Answer:

5m/s, INELASTIC COLLISION

Discussion
50.

The moment of force about a fixed in a of a rod ab of length 1.5 m 60 and. what is the force applied at the free point B of the rod? Also draw the diagram.Only the diagram is important.whom will answer this question I will mark him/her as brainliest answer and I will also follow him/her.​

Answer»

ANSWER:

I didn't KNOW the answer ☺️ SORRY
Discussion