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The average speed of an object is defined as the total path length traveled divided by the total time elapsed

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√2gh will be the velocity of the object before reaching the surface of the earth

As we know in SHM100=16

Omega,hence Omega=100\16(from given value)

Now,Again ,5=100/16√A(square)-X(square)

Hence X=3.37cm from mean position

Given, Amplitude , A = 4cm velocity of the particle at mean position , v₀ = 10cm/s We know the relation between speed of particle , amplitude and distance of particle from mean position is given by v = ω√{A² - x²} Here ω is angular frequency,

speed at mean position , v₀ = ωA = 10 cm/s speed at x distance from mean position , v = ω√{A² - x²} = 5 cm/s So, ωA/ω√{A² - x²} = 10/5 ⇒ A/√{A² - x²} = 2 ⇒ 4/√(4² - x²) = 2 ⇒ 4 = 4² - x²⇒ x² = 12 ⇒ x = ±2√3 cm

Hence, position of particle where speed = 5cm/s is ±2√3 cm

Escape velocity of a projectile from the Earth,vesc= 11.2 km/sProjection velocity of the projectile,vp= 3vescMass of the projectile =mVelocity of the projectile far away from the Earth =vfTotal energy of the projectile on the Earth = (1/2)mvp^2– (1/2)mvesc^2Gravitational potential energy of the projectile far away from the Earth is zero.Total energy of the projectile far away from the Earth = (1/2)mvf^2From the law of conservation of energy, we have(1/2)mvp2– (1/2)mvesc^2 = (1/2)mvf2vf= (vp^2–vesc^2)^1/2= [ (3vesc)^2–vesc^2]^1/2= √8vesc= √8×11.2 = 31.68 km/s.

average velocity will be zeroas he returned back to the original position hence the displacement is zero

average speed =total distance/total time=24+24/4+12=48/16=3m/sec

Given,

u = 20m/s

θ = 45°

Maximum height in projectile is given by,

h = u²sin²θ/2g

= (20)²sin²45°/2*9.8

= 100/9.8

= 10.204 m

Thus, maximum height attained by the ball is 10.204 m.

Mass of rocket(M)= 5000 kg

Exhaust speed(v)=800m/s^2

Acceleration(a)=20m/s^2

acceleration due to gravity (g)=10 m/s^2

F=m(g+a)=5000(10+20)=15000N

amount of gas ejected per second= F/v=15000/800=187.5 kg/s

Power=energy/timeEnergy(j) = work done (j)Hence , power = workdone / timeWorkdone =force[the weight-mg] X distanceWorkdone = 75(10)x(20x10)=150000Thus , power = 150000/5s=30000watts

fuck off vghhbbbbbbbbhgffhyffccvhhgfxcvj

X = a + bt^2v = 2bt

Velocity at t = 0 is 0Velocity at t = 2s is = 2*2.5*2 = 10 m/s

At t = 2sx = 8.5 + 2.5*2^2x = 18.5m

At t = 4s x = 8.5 + 2.5*4^2x = 48.5m

Average velocity = (48.5 - 18.5) / 2 = 18m/s

sis exam point telling the answer

sis exam point telling the answer

When the object travels in a circular path it's average speed is not zero but it's average velocity is zero because it's Displacement will also be zero..

Average velocity = Displacement/time

So,the average velocity will also be zero.

when any body or an object travels in a circular path it's average speed is not zero but it's average velocity is zero.We know that -Average velocity = Displacement / time So , average velocity will be zero.

circular motion in which the average speed is not zero but average velocity is zero

X = a + bt^2v = 2bt

Velocity at t = 0 is 0Velocity at t = 2s is = 2*2.5*2 = 10 m/s

At t = 2sx = 8.5 + 2.5*2^2x = 18.5m

At t = 4s x = 8.5 + 2.5*4^2x = 48.5m

Average velocity = (48.5 - 18.5) / 2 = 18m/s

A body os said to have uniform velocity if it covers equal distance in equal interval of time in a particular direction, however the time intervals may be small.This means that if a body is travelling at a constant speed along a straight line then it is said to have uniform velocity.

i)A body is said to haveuniform velocityif it covers equal distance in equal intervals of time in a particular direction, however the time intervals may be small. Thismeansthat if a body is traveling at a constant speed along a straight line (i.e, a particular direction) then it said to haveuniform velocity.

ii)Theaverage velocity can bedefinedas the displacement divided by the time.

iii)Instantaneous velocityis thevelocityof an object in motion at a specific point in time. This is determined similarly to averagevelocity, but we narrow the period of time so that it approaches zero. If an object has a standardvelocityover a period of time, its average andinstantaneous velocitiesmay be the same.

Write S'l unit of speed and velocity

define resultant velocity

The average velocity of an object is its total displacement divided by the total time taken. In other words, it is the rate at which an object changes its position from one place to another. Average velocity is a Vector quantity. The SI unit is meters per second.

Instantaneous velocity is defined as the rate of change of position for a time interval which is very small (almost zero).

Average velocity is equal to the instantaneous velocity when acceleration is zero. In order for acceleration of an object to equal zero, there can be no change in speed or direction. For example, when a car is traveling down a straight road on flat land using cruise control.

Let assume the distance be 2X.

By the question ,

AVERAGE VELOCITY (V)= TOTAL DISPLACEMENT / TIME TAKEN

Total displacement = 2X

Have to find time taken for the V1 and V2.

time taken for the first half = X/V1

time taken for the next half = X/V2.

substitute value in AVERAGE VELOCITY

V =2S/(X/V1 + X/V2)

take "x" common and take L.C.M

V= 2S /X( V1 + V2 )/V1 V2

V= 2S (V1 V2)/ X (V1 + V2).

Let Let T be the time taken by the person to cover complete distance.

Case I:let x1 be the half distance covered. speed=v1time taken to cover the first half of the journey :t1=distance/speed=(x/2)/v1

Case IIlet x2 be the half distance covered.speed=v2time taken to cover the second half of the journey :t2=distance/speed=(x/2)/v2

Total time=T=t1+t2=x/(2v1)+x/(2v2)

T=x/2[1/v1+1/v2]

=x[v1+v2]/2v1v2

Average Velocity= Displacement/ total time=x/x[v1+v2]/2v1v2=2v1v2/(v1+v2)

∴Average velocity is given by 2v1v2/(v1+v2)

you are correct i was just testing the app

but brainly is much better than this app

Here we have to check the dimensions of 1/2(m)v² and mgh

foe 1/2mv² the dimension is M(L/T)² = ML²T-²

now for mgh , the dimension is M(L/T²)L = ML²T-².

so we can say that the equation is dimensionally correct

..Induced EMF= (change in Magnetic Flux Density x Area)/change in Time.induces emf is opposite to the emf appliedhenceoption b

Let the time at which the Bus A overtakes the bus B be T seconds.

∴ Distance travelled by the Bus A from the Origin in Time T = Speed× Time= 10× T= 10 TTotal Distance of Bus A = 50 + 10T

Now,Distance travelled by the Bus B in time T seconds = 5× T= 5TTotal Distance of Bus B = 100 + 5T

We know,Distance of Bus A after T seconds = Distance of the Bus B after T seconds.

⇒ 50 + 10T = 100 + 5T⇒ 10T - 5T = 100 - 50⇒ 5T = 50⇒ T = 50/5⇒ T = 10 seconds.

Time at which the Bus A overtake the Bus B is 10 seconds.

Now, Total Distance travelled by the Bus A from the origin = 50 + 10T= 50 + 10(10)= 50 + 100=150 m.

∴ Distance from the origin at which the Bus A overtake the Bus B is 150 m.

Along vertical = v = u + at= vSinθ’ = uSinθ - gt= v Sin 30 = 30 * u Sin 30 + 1 * 10= v * 1/2 = 30 * 1/2 + 10= v / 2 = 25= v = 5 m/s.

distance travelled = velocity× time = 30*30 = 900m

Using relation

Vf - Vi = atVf - 10 = 5 * tt = (Vf - 10)/5

Now,S = Vi*t + 1/2 at^230 = 10*(Vf - 10)/5 + 1/2*5*[(Vf-10)/5]^2

30 = 2Vf - 20 + 5/2 *(Vf - 10)^2/25

30 = 2Vf - 20 + 5/2(Vf^2 + 100 - 20Vf)/25

30 = 2Vf - 20 + 1/10(Vf^2 + 100 - 20Vf)

50*10 = 20Vf + Vf^2 + 100 - 20Vf

Vf^2 = 500 - 100 = 400

Vf = 20

Therefore,Final Velocity = 20 m/s

thanks a bunchby the way are you really a grade 8 student?

F= 2000hzw.l. = .35m

speed = f × w.l.700m/s

now dist. = 1500 mtime= d/sp.1500/70015/7s2.142sec.

Since the normal makes an angle of 90 with the plain.Therefore,Let angle between incident ray and plain be X.=> angle of incidence=(90-X)=> angle of refraction=(90-X)According to question:90-x+ 90-x=130=>180-2x=130=>2x=50=x=25.

Here we are given the frequency as 4 kHz

We know that 1 kHz = 1000 Hz

So , 4kHz = 4000 Hz

Frequency, f = 4000 Hz

Also ,

Wavelength , λ = 40 cm

1 cm = 0.01 metres

So , Wavelength , λ = 0.40 metres

We know that 1 kilometre = 1000 metres

So ,3.2 kilometre = 3200 metres

We know that ,Velocity = frequency x Wavelength

= 4000 x 0.40

= 1600 m/s

Also ,

Time = Distance/Velocity

= 3200 / 1600

= 2 s

The opposing force experienced when one surface rolls over the other is called rolling friction.

rolling frictionorrollingdrag,isthe force resisting the motion when a body (such as a ball, tire, or wheel) rolls on a surface. .

Rolling friction is the force of resistance by the opposite objects.

1 limiting friction .... when it got equals to sign ....

limiting friction... you got to have a better concentration on the graph... you can very well see that limiting friction will have the peak while other don't. Limiting friction is the maximum value of static friction.

limiting friction is the max value of static friction.

1) limiting friction

1)'' Limiting friction'' the maximum value of static friction.So, option 1st is correct answer.

the maximum friction of static friction is limiting friction

limiting friction is right answer

the option 1st is right

maximum value of static friction is limiting friction

limiting friction has the max. value

limiting friction is the answet

limiting friction is the maximum value of static friction

1) limiting friction is the max value of static friction therefore the correct answer is the limiting friction

Thesliding friction is less than static frictionbecause of the interlocking of irregularities in the two surfaces. When the object startsslidingthe contact points on its surface, do not get enough time to lock into the contact points on the floor. So thesliding frictionis slightlyless thanthestatic friction.