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The light body.

Kinetic energy equation:Ek=1/2mv2Ek=1/2mv2Momentum equation:p=mvp=mvKinetic energy equation in terms of momentum:Ek=p^2/2mEk=p^2/2m

If two objects have the same mass and the same momentum, they have the same kinetic energy.Of a light and a heavy body having equal kinetic energy, the heavy body has more momentum.At equal momentum, the light body has more kinetic energy.

R=pl/aLet, L=2l A=a/2 (for keeping same volume), R=pL/A 15=pL/AWe have to find, r=pl/aSo, 15=pL/A 15=p2l/(a/2)

15=p 2l2/a15=p 4l/a15/4 = pl/aSo, r=15/4

Whilekinetic energyis theenergywhich an object contains because of a particular motion. On the other hand,potential energyis the storedenergy, because of its state of rest.Role of latent heat in state of change of substance :Latent heatdoes not raise the temperature.But thelatent heat has always to be supplied to changethestate of a substance.It is calledlatent heat because it becomes hidden in the substanceundergoing thechange of state,and does not show its presence by raising the temperature.

i) ester. -COO- ii) ether -O-iii) C.A -COOHiv). ketonee. -CO-

Young’s modulus Y = Force x L / A x l

Or rewriting we have Y = WL/Al

\l= WL/ AY

Since the length of the wire is on both sides of the pulley length L/2 will be on both sides.

Thus the elongation of the wire in mm is l mm.

hence answer is 2mm

an object that sinks displaces an amount of fluid equal to the object's volume. Thus buoyancy is expressed through Archimedes' principle, which states that the weight of the object is reduced by its volume multiplied by the density of the fluid.

1. equivalent resistance = 13.5 ΩJoin two resistors in parallel and one in series as shown in 1st figure We know, in parallel , 1/R' = 1/R₁ + 1/R₂ Here , R₁ = R₂ = R₃ = 9Ω Now, 1/R' = 1/9Ω + 1/9Ω ⇒ R' = 9/2Ω = 4.5 Ω So, equivalent resistance , Req = R' + R₃ [ ∵ R' and R₃ in series ] ∴ Req = 4.5Ω + 9Ω = 13.5Ω

2. Equivalent resistance = 6Ω two resistors are in series which is parallel with 3rd resistor as shown in 2nd figure .now, R' = R₁ + R₂ = 9Ω + 9Ω = 18Ω [ R₁ and R₂ are in series ] now, equivalent resistance, Req = R'R₃/(R' + R₃) [ R' and R₃ are in parallel ] ∴ Req = 18 × 9/(18 + 9) = 18 × 9/27 = 6Ω

two resistor in parallel and third resistor in series

Diamagneticmaterials have a weak, negative susceptibility to magnetic fields. Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom. Diamagnetic properties arise from the realignment of the electron paths under the influence of an external magnetic field. Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.

Paramagneticmaterials have a small, positive susceptibility to magnetic fields. These materials are slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field. Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.

Ferromagneticmaterials have a large, positive susceptibility to an external magnetic field. They exhibit a strong attraction to magnetic fields and are able to retain their magnetic properties after the external field has been removed. Ferromagnetic materials have some unpaired electrons so their atoms have a net magnetic moment. They get their strong magnetic properties due to the presence of magnetic domains. In these domains, large numbers of atom's moments (1012to 1015) are aligned parallel so that the magnetic force within the domain is strong. When a ferromagnetic material is in the unmagnitized state, the domains are nearly randomly organized and the net magnetic field for the part as a whole is zero. When a magnetizing force is applied, the domains become aligned to produce a strong magnetic field within the part. Iron, nickel, and cobalt are examples of ferromagnetic materials. Components with these materials are commonly inspected using the magnetic particle method.

Diamagnetic Material

A material that turns at a right angle to the field by producing a magnetic response opposite to the applied field is called diamagnetic material such as silver, copper, and carbon have permeability’s slightly less than free space (for copper, μr= 0.9999980).

Paramagnetic Material

A material aligning itself with the applied field is called paramagnetic material. Paramagnetic materials such as aluminum and air have permeability’s slightly greater than that of free space (for air μr=1.0000004). The effects of diamagnetic and paramagnetic are negligibly small so that materials possessing these weak phenomena are said to b non-magnetic.

Ferromagnetic Material

Within the paramagnetic class of materials, is a special classification of materials called ferromagnetic material. These materials are strongly attracted to magnets and exhibit Paramagnetism to a phenomenal degree. Ferromagnetic materials such as iron, steel, cobalt and their alloys have relative permeability’s extending into the hundreds and thousands, are said to be magnetic.

The magnetic properties of matter are associated with the spinning motion of electrons in the third shell of the atomic structure.

Paramagnetismis a form of magnetism whereby some materials are weakly attracted by an externally applied magnetic field, and form internal, induced magnetic fields in the direction of the applied magnetic field. ... Due to their spin, unpaired electrons have a magnetic dipole moment and act like tiny magnets.

Let’s say that the bike starts from zero and accelerates at 5m/s^2 for t secs until it reaches v m/s

Then (v-0)/t = 5 or v=5t , and -v = -5t

If T is the total time then there is T-t secs left to stop using a deceleration of

-10 m/s^2 ,so

(0-v)/(T-t) =-10

-v = -10T +10t but -v = -5t , so

-5t = -10T + 10t

10T = 15t

T = 3t/2

Now use the piece of information given the distance 1500m

[1/2] v T = 1500

v T =3000 . Note we have v and T in term of one [1] unknown t

[5t][3t/2]=3000

15t^2 = 6000

t^2 = 400

t=20

So T = 3t/2= 3[20]/2 = 30 secs is the shortest time to complete the journey given the conditions.

therefore answer is 30secs

2+1 ll 2+13*3/3+3=1.51.5 ll?1.5*3/1.5+3=1

3 ohm is correctStups are given👆👆

The maximum speed of the particle corresponds to the area of the a→t graph.

Maximum speed = Area of the graph

= 1/2 ×10 × 11

= 55 m/s

U = sqrt(2gH)= sqrt(2 × 9.8 × 0.25)

v = sqrt(2gh)= sqrt(2 × 9.8 × 0.09)

e = v / u= [sqrt(2 × 9.8 × 0.09)] / [sqrt(2 × 9.8 × 0.25)]= 0.6

Coefficient of restitution is 0.6

Givenu = 20 km/ hv = 10 km/htime = 3 min. convert in hr. 3/60 =0.05 hr.solutionA =

A = v - u / t10 - 20 / 0.05 = -200 km/h

Yes, the statement is correct. A convex lens of focal length 20 cm will produce a magnified virtual image if object is placed at a distance less than 20 cm from the lens.A convex lens of focal length 20 cm will produce a magnified real image if object is placed at a distance greater than 20 cm and less than 40 cm from the lens

Work is said to be done when force applied on an object shows the displacement in that object.

W = F x s.

Aforceacting on an object may cause the object to change shape, to start moving, to stop moving, to accelerate or decelerate. When two objects interact with each other they exert aforceon each other, the forcesare equal in size but opposite in direction.

F=18O=x cm (let)I= x/2 cmwe have,I/O=v/uor, x/2x= v/uv= u/2again,1/f=1/u+1/v1/18= 1/u+2/uu=54cm Ans#

Workis said to bedonewhen an object moves (displaces) along the direction of application offorce." OR Theworkis defined asforcedisplacement. W = F x S

Types of workPositive workWhen a spring is stretched, the work done by the stretching force is positive as the force acts in the direction of displacement of the spring.

negative work The work done against friction is negative as the frictional force is always in direction opposite to displacement.

zero workA person carrying a load on his hand and walking on a horizontal road does no work as the force is acting at right angle to displacement.

Work done by a force.Workisdonewhenever aforcemoves something over a distance. You can calculate the energy transferred, orwork done, by multiplying theforceby the distance moved in the direction of theforce. ... The transfer of energy is not 100% efficient and not all the energy transferred is represented by mgh

Special cases:

CaseI: When the body moves in the direction of the force. In this case, q = 0o.

from eq (i), we have

W = FScosoo

i.e. W = FS …(ii)

i.e. W = product of the force and the distance through which the body moves in the direction of the force.

This type of work is said to be positive. Therefore, work done by a force is positive is q < 90o

Examples of Positive Work:

(1) When a spring is stretched, the work done by the stretching force is positive as the force acts in the direction of displacement of the spring.

(2) When a body falls freely under gravity, the work done by gravity is positive as force (gravity) and displacement of the body are in the same direction.

(3) When a gas taken in a cylinder expands, work done by the force is positive as the displacement is in the direction of the force.

Case II:. When the force acting on a body and its displacement are in the opposite direction

In this case, q = 180o. Therefore, from eqn (i), we have

W = FS cos180o= FS(-1)

Or W = -FS …. (iii). This type of work is said to be negative. Therefore, work done by a force is negative if q > 90o.

Examples of Negative Work:

(i) The work done against friction is negative as the frictional force is always in direction opposite to displacement.

(ii) Work done in lifting a body is negative as gravitational force and displacement are in the opposite direction.

Case III:. When force and displacement are at right angle to each other.

In This case q = 90o

From eqn (i), we have

W = FS cos90o= FS . 0 = 0

Therefore, no work is done by a force which acts at right angle to the direction of displacement of the body.

Examples of Zero Work:

(i) The work done by the centripetal force acting on a body moving in a circle with uniform speed is zero as it is always at right angle to the direction of motion of the body.

W = FS cos90o

= FS (0)

= 0

(ii) A person carrying a load on his hand and walking on a horizontal road does no work as the force is acting at right angle to displacement.

Case IV: If S = 0, then from eqn. (i), we have

W = f x 0 x cosq = 0

Therefore, no work is done by because if it can not displace a body on which it acts.

Types of Cooling Towers1.Mechanical draft cooling tower.2.Atmospheric cooling tower.3.Hybrid draft cooling tower.4.Air flow-characterised cooling tower.5.Construction-characterised cooling tower.6.Shape characterized cooling tower.7.Cooling tower based on method of heat transferEach of these could carry several cooling towers. For instance, categorizing cooling towers in terms of heat transfer method gives three options: Dry cooling towers, Open circuit cooling towers and Closed circuit cooling towers/ fluid cooling towers.

It is the process in which metals are slowly eaten up by the action of air moisture or chemicals. For example rusting is a form of corrosion in which iron is eaten up by the action of air and moisture and a reddish brown coating of iron oxide is formed as shown in the following chemical reaction.

1. Galvanic corrosionis the most common and impactful form of corrosion. It occurs when two dissimilar (different) metals are in contact in the presence of an electrolyte. In a galvanic cell (bimetallic couple), the more active metal (anode) corrodes and the more noble metal (cathode) is protected. There are a number of factors that affect the galvanic corrosion including types of metals, relative size of anode, and environment (temperature, humidity, salinity, etc.)

2. Microbial corrosion, commonly referred to as microbiologically influenced corrosion (MIC) is caused by microorganisms. It applies to both metallic and non-metallic materials with or without oxygen. When oxygen is absent, sulfate-reducing bacteria are active and produce hydrogen sulfide causing sulfide stress cracking. When oxygen is present, some bacteria may directly oxidize iron to iron oxides and hydroxides. Concentration cells can form in the deposits of corrosion products, leading to localized corrosion.

pls mark as the best

Unlike other planets inthe solarsystem that rotatewesttoeast(the same spinning direction ofthe Sunor prograde rotation),Venusspinseasttowest, (opposite tothe Sunor retrograde rotation). Hencethe sun riseflips in direction.

Distance to be covered by the parrot=150m

speed of train=10m/s

speed of parrot=5m/s

now consider the train with respect to parrot and not with respect to land.

velocity of parrot=5

velocity of train=-10(∵the train is in opposite direction)

so relative velocity of train w r t parrot = 5 - (-10)=15m/s

now time = distance/speed = 150/15= 10s

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75 Centimeter Mercury (0°C) (cmHg)

=

99991.77632 Pascal (Pa

the distance travelled by train=50+100+200+400=750kmthe magnitude of displacement = 400km

Magnetic Tape