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Bending of a cyclist in circular motion : 1. A cyclist while going round a curve on a horizontal track has to bend himself a little from his vertical position in order to avoided overturning. When he bends himself inward, a component of the reaction of road provides him the necessary centripetal force for circular motion.2. In order to find the angle through which a cyclist has to bend himself to take a safe turn, let us assume thatm= combined mass of cycle and cyclistv= uniform speed with which the cyclist takes the turnr=radius of circular trackAnd θ = angle through which the cyclist bend from vertical3. Suppose G represents the centre of gravity of the system . If force of friction between the cycle and the road is not taken into account, then following two forces act on the system at the point G.a) Weight mg of the system acting vertically downloads through the centre of the gravity of the system.b) Reaction R exerted by the road on the system acting along GR at the angle θ with the vertical.4. The reaction R can be resolved into two rectangular components.a) R cosθ, acting vertically upwards,b) R sinθ, acting horizontally i.e, along the radius and towards the centre of the circular track horizontal component R sin θ provides the necessary centripetal force. Hence:R cos θ=mg………………(1)Rsin θ = mv2/r ………..(2)Dividing eq.(2) by eq. (1), we haveR sin θ/R cos θ=(mv2/r)/mg or tan θ v2/rgOrθ = tan-1(v2/rg)

Theentireplanet isn’t blue, of course. The clouds themselves are white, reflecting the white, direct sunlight back out at any onlookers. Ice — such as the caps on our planet’s poles — appears white for the same reasons. The continents, similarly, appear either brown or green from a great distance, depending on the seasons and how covered-in-plants the terrain isThis teaches us something important: the Earth isn’t blue because the sky/atmosphere is blue. If that was the case, all the light reflecting off of the surface would be blue-hued, and we simply don’t see that. But there’s a hint that comes from the truly blue parts: Earth’s seas and oceans. Theshadeof blue that the Earth’s water appears varies based on how deep the water is. You’ll notice, if you look closely at a photo like the one below, that the watery regions borderingthe continents (along the continental shelves) is a lighter, more cyan shade of blue than the deep, dark depths of the ocean.

change in momentum is of 40kgm/s

We know kinetic energy = 1/2 ×m× v ²SI unit of mass = kilogramsSI unit of velocity = m/sso dimensional formula of K.E= [ L²M¹T^-2]

Speed of Bus ASpeedA = Distance /Time = 360/5 = 72 km/hr

Speed of Bus BSpeedB = Distance /Time = 426/7 = 60.8 km/hr

SpeedA > SpeedBThus Bus A is faster than Bus B

C. Lipase is correct option

Ohm is the SI unit of electrical resistance.

1ohm is equivalent to

According to the Newton’s 2ndLaw of motion, the rate of change of linear momentum of a body is directly proportional to the applied external force and in the direction of force.

It means that the linear momentum will change faster when a bigger force is applied.

Consider a body of mass ‘m’ moving with velocity v.

The linear momentum of a body is given by:

p = mv

Now According to Newton’s 2ndLaw of Motion:

Force is directly proportional to rate of change of momnetum, that is

F α dp/dt

F = k dp/dt

F = k d(mv)/dt

F = k md(v)/dt

F = k ma

Experimentally k =1

F = k ma

Which is the required equationof force.

the liquid metal MERCURY

the liquid metal is MERCURY

mercury is the answer of this question

mercury is the answer

the metal.which is liquid at room temperature is mercury

mercury is the answer of this question

Mercury is the answer

it is mercury which is liquid at room temperature

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Mercury is the metal which is liquid in room temperature

the answer is mercury

Mercury is the right answer to this question

mercury is the answer of this question which is found in liquid State

mercury is a very good app for the

The answer is liquid mercury

mercury is a very good source of energy

the liquid metal is mercury

Power of each bulb=50W=0.05 KWpower of both the bulbs=2*0.05=0.1 KW

units used by the bulbs in a day=0.1*5=0.5 units

power of each fan=70W=0.07KWpower consumed by both fans=0.07*2=0.14units used by the fans in a day=0.14*10=1.4 unitstotal units consumed in a day=0.5+1.4=1.9 unitsunits of energy consumed in a month of 30 days=30*1.9=57 unitsans. therefore 57 units of electricity is consumed ina month in the house

(i) The cornea acts as the eye's outermost lens. It functions like a window that controls and focuses the entry of light into the eye. The cornea contributes between 65-75 percent of the eye's total focusing power. When light strikes the cornea, it bends -- or refracts - the incoming light onto the lens.

(ii) The iris is a thin, circular structure in the eye, responsible for controlling the diameter and size of the pupil and thus the amount of light reaching the retina. Eye color is defined by that of the iris.

(iii) The whole job of pupil is to control the amount of light that gets into the eye. It's called a pupillary reflex, and you have probably noticed that a person's pupils are smaller in bright light and bigger in low light.

(iv) The retina is a thin layer of tissue that lines the back of the eye on the inside. It is located near the optic nerve. The purpose of the retina is to receive light that the lens has focused, convert the light into neural signals, and send these signals on to the brain for visual recognition.

The correct answer should be - 200 J of energy.

Going by the hierarchy of the Trophic levels, the plants are primary producers, the herbivores are the primary consumers, and the carnivores are the secondary consumers.

So, according to the 10% law, at every trophic level, the energy is lost in the food due to biological processes like digestion, or energy lost during transfer or during the breakdown of respiration and other functions. Only 10% of the energy is stored in the flesh or in the body parts.

Therefore according to the 10% rule, if the plants have 20000 J of energy, then when it comes down to next trophic level of the herbivores, the energy level will come down to 2000 J. When it comes down to the nexttrophic level of the carnivores that includes Man, the energy available will be only 200 J.

Here N1 is the contact force by incline and N2 that by the vertical wall.

l

animal and plant waste can be converted into manure, which can be used for plantation and farming

latent heat is the latent heat of fusion.

1/sin theta = cosec theta. so, Option A

De Broglie first used Einstein's famous equation relating matter and energy,

E=mc2,E=mc2,

whereE=E=energy,m=m=mass,c=c=speed of light.

Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation,

E=hν,E=hν,

whereE=E=energy,h=h=Plank's constant (6.62607×10−34Js6.62607×10−34Js),ν=ν=frequency.

Since de Broglie believes particles and wave have the same traits, the two energies would be the same:

mc2=hν.mc2=hν.

Because real particles do not travel at the speed of light, de Broglie substitutedvv, velocity, forcc, the speed of light:

mv2=hν.mv2=hν.

I want a direct proof without substitutingvvforcc. Is it possible to prove directlyλ=h/mvλ=h/mvwithout substitutingvvforccin the equationλ=h/mcλ=h/mc?

good answer right away and

Pole Star is a star which appears stationary from the Earth .

On a clear night , go to an open ground . Choose the time of the year when a group of stars , called the Saptrishi or Ursa Major , is prominent in the sky . Look towards the northern part of the sky and identify Ursa Major .

Imagine a straight line passing through these stars . Extend this imaginary line towards the north direction ( nearly five times the distance between the two stars ) . This line will lead to a star , known as the Pole Star ( North Star )

As there is no scattering of sunlight in space, it appears dark

The sky appears dark instead of blue to an astronaut because there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reach the eyes of the astronauts and the sky appears black to them.

Thesky appears dark instead of blue to an astronautbecause there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reach the eyes of theastronautsand thesky appears blackto them.

there is no atmosphere in outer space to light can scattered the sunlight. as there is no scattered ,no light reach to eyes of astronaut and sky appears black.

The sky appears dark instead of blue to an astronaut because there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reach the eyes of the astronauts and the sky appears black to them.

sky appears dark instead of blue to an astronautbecause there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reach the eyes of theastronautsand thesky appears blackto them.

sky appears dark instead of blue to an astronaut because there is no atmosphere in outer space that can scatter that has a solid is not connected no light reach the eyes of the astronaut and so the sky looks dark.

The sky appears dark instead of blue to an astronaut because there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reach the eyes of the astronauts and the sky appears black to them.

Thesky appears dark instead of blue to an astronautbecause there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reach the eyes of theastronautsand thesky appears blackto them.

A clear cloudless day-time sky is blue because molecules in the air scatter blue light from thesunmore than they scatter red light. When we look towards thesunat sunset, we see red and orange colours because the blue light has been scattered out and away from the line of sight

Option (a) is correct.

thank you so much

Answers-T = 1384 sR = 27712 mTa = 692 s Explaination------------------------Given-u = 800m/sθ = 60°

Solution-a) Time of flight-T = 2usinθ/gT = 2×800×sin60T = 1384 s

b) Horizontal range,R = u^2sin2θ/2gR = 800^2 × sin120 / (2×10)R = 32000×0.866R = 27712 mR = 27 km

c) Time of ascent-Ta = usinθ/gTa = 800×sin60Ta = 692 s

Given, V = 50v

Therefore,

E=qV=1.6×10^−19×50=8×10^−18J

We know, E=p^2/2m

8×10^−18×2×1.673×10−27=p^2

h/λ=(267.68×10−46)^1/2

h/λ= 16.36×10−23

λ=6.62×10^−34/16.36×10^−23

λ=4.04×10^-10 m

Wavelength of proton = 4.04×10^-10 m

V=144Vwavelength=12.27/√VV=144hence wavelength=12.27/12=1.2A

height of aeroplane is approx.=233mhe remains in the air for approximately =15s

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Let the inital distance traveleld before opening parachute be h1 = 40m

then s = ut + (1/2)gt2

as u =0

40= (1/2)g.(t1)2

or

t1 = √80/9.8 = √400/49 = 20/7 s

his velocity at that time is

v1 = u + gt1 = 10 X 20/7 = 28.28 m/s

Then there is retardation of 2 m/s

v22- v12= 2gh2

so

32= (28.28)2- 2g.h2

or h2 = 25x31/(2x9.8)

Total height will be h = h1 + h2 = h2+40

For calculating the time after parachute opens,

v2 = v1 - at

3 = 28 - 2t2,

or

t2=25/2

Total time, t = t1 + t2 = 25/2+20/7

ortotal time of flight would be t = 15.35 s

tanslatory : rotational kinetic energy is 5:2Ans:(b)

rotational K.E = 1/2*(I)(w²)

and anglular momentum is = Iw

so, here K.E is same

=> 9(w1²) = 1(w2²)=> w1/w2 = √1/9 = 1/3..

so, Angular momentum ratio will be = (I1)(w1)/(I2)(w2) = 9*w1/(1)*(3w1) = 3/1

so, the value is 3:1

option 1.