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It is BASED on ARCHIMEDES principle
two applications are
1.submarines and ships are designed on this principle
1.hydrometer is used to CHECK the DENSITY of LIQUID is based on this principle

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⏩ Here's u'r answer .....⏪

*⃣of the earth to be zero. Let us see an example of a FRUIT falling from a tree.

Consider a point A, which is at some height ‘H’ from the ground on the tree, the velocity of the fruit is zero hence potential energy is maximum there.

E = mgH ———- (1)

When the fruit is falling, its potential energy is DECREASING and kinetic energy is increasing.

At point B, which is near the bottom of the tree, the fruit is falling freely under gravity and is at a height X from the ground, and it has a speed as it REACHES point B. So, at this point it will have both kinetic and potential energy.

E = K.E + P.E

P.E = mgX ——— (2)

According to third equation of motion,

v2=2g(H–X)⇒12mv2=12m.2g(H–X)⇒K.E=12m.2g(H–X)⇒K.E=MG(H–X)

K.E=mg(H-X)——– (3)

Using (1), (2) and (3)

E = mg(H – X) + mgX

E = mg(H – X + X)

E = mgH

Similarly, if we see the energy at point C which is at the bottom of the tree, it will come out to be mgH. We can see as the fruit is falling to the bottom and here, potential energy is getting converted into kinetic energy. So there must be a point where kinetic energy becomes equal to potential energy. Suppose we need to find that height ‘x’ from the ground. We know that at that point,

K.E = P.E

=> P.E = K.E = E2 ——– (4)

E2 is the new energy

Where, E = mgH2

H2 is the new height.

As the body is at height X from the ground,

P.E = mgX ——— (5)

Using (4) and (5) we get,

mgX=mgH2⇒X=H2

H2 is referred to the new height......*⃣


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Do it by yourself
Do it by yourself
Do it by yourself
Do not DISTURBANCE to OTHERS

But why do you WANT ..........

B.) A is the ANSWER to the question.
Hope it helps you .
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Derivation of E=mc2

There are quite a few ways to derive Einstein’s famous equation E=mc^{2}. I am going to show you what I consider to be the simplest way.  Feel FREE to comment if you think you know of an easier way.

We will start off with the relationship between ENERGY, force and distance. We can write

dE = F dx \text{ (1) }

Where dE is the change in energy, F is the force and dx is the distance through which the OBJECT moves under that force.  But, force can also be written as the rate of change of momentum,

F = \frac{dp}{dt}

Allowing us to re-write Equation (1) as

dE = \frac{dp}{dt}dx \rightarrow dE = dp \frac{dx}{dt} = vdp \text{ (2) }

Remember that momentum p is defined as

p =mv

In classical physics, mass is constant. But this is not the case in Special Relativity, where mass is a function of velocity (so-called relativistic mass).

m = \frac{ m_{0} }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } \text{ (3) }

where m_{0} is defined as the rest mass (the mass of an object as measured in a reference frame where it is stationary).

Assuming that both m \text{ and } v can change, we can therefore write

dp =mdv + vdm

This allows us to write Equ. (2) as

dE = vdp = v(mdv + vdm) = mvdv + v^{2}dm \text{ (4) }

Differentiating Equ. (3) with respect to velocity we get

\frac{dm}{DV} = \frac{d}{dv} \left( \frac{ m_{0} }{ \sqrt{ (1 - v^{2}/c^{2}) } } \right) = m_{0} \frac{d}{dv} (1 - v^{2}/c^{2})^{-1/2}  

Using the chain rule to differentiate this, we have

\frac{dm}{dv} = m_{0} \cdot - \frac{1}{2} (1 - v^{2}/c^{2})^{-3/2} \cdot (-2v/c^{2}) = m_{0}  (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-3/2} \text{ (5) }

But, we can write

(1 - v^{2}/c^{2})^{-3/2} as (1-v^{2}/c^{2})^{-1/2} \cdot (1-v^{2}/c^{2})^{-1}

This allows us to write Equ. (5) as

\frac{dm}{dv} = m_{0}  (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-1} \cdot (1 - v^{2}/c^{2})^{-1/2}  

From the definition of the relativistic mass in Equ. (3), we can rewrite this as

\frac{dm}{dv} = \frac{ m v }{ c^{2} }(1-v^{2}/c^{2})^{-1}

Which is

\frac{dm}{dv} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}} - \frac{ v^{2}}{c^{2} } \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}-v^{2}}{c^{2}} \right)^{-1}  = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}-v^{2}}   \right)  

\frac{dm}{dv} = \frac{ m v }{ (c^{2}-v^{2}) } \text{ (6) }

So we can write

c^{2}dm - v^{2}dm = mvdv  

Substituting this expression for mvdv into Equ. (4) we have

dE = vdp = vd(mv) = mvdv + v^{2}dm = c^{2}dm - v^{2}dm + v^{2}dm  

So

dE = c^{2} dm  

Integrating this we get

\int_{E_{0}}^{E} dE = c^{2} \int_{m_{0}}^{m} dm  

So

E - E_{0} = c^{2} ( m - m_{0} ) = mc^{2} - m_{0}c^{2}  

E - E_{0} = mc^{2} - m_{0}c^{2}  

This tells us that an object has rest mass energy E_{0} = m_{0}c^{2} and that its total energy is given by

\boxed{ E = mc^{2} }  

where m is the relativistic mass.

I don't KNOW srrryyyy

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*⃣ OBTAINING an image on a screen : it may MEANS having a an image on SURFACE *⃣

Cysts are usually noncancerous and have a sac-like structure that can contain FLUID, pus, or gas. Cysts are common and can occur anywhere on the body. Cysts are often CAUSED by infection, clogging of sebaceous glands, or around earrings.
Many cysts do not cause any symptoms and go away on their own. Cysts can come back. Draining or surgically removing cysts usually has no complications or side effects. In RARE cases in which a cyst is next to or inside a cancerous tissue, the prognosis depends on the type of cancer and whether it has spread.

In simple language you can say that :-
a SWELLING or a lump filled with liquid in the body or under the skin is known as cyst.

hope it is helpful to U... :-)

It is KNOWN as heating EFFECT .
HOPE it helps you.
Please mark the answer as the BRAINLIEST one.

Since the collision is PERFECTLY inelastic, the NET linear MOMENTUM will be conserved.

So M1V1+M2V2=(M1+M2)V
Or (7)(1) + (2)(-2)=(3)V
Or V=1m/s

So, LOSS in kinetic energy=1/2M1(V1)^2 + 1/2M2(V2)^2 - [1/2(M1+M2)V^2]
= 1/2(1)(7^2) + 1/2(2)(2^2) - 1/2(3)(1^2)
=49/2 + 4 - 3/2
=27J

Pressure=Force/Area

=10/1=10 Newton/meter²

=10 PASCAL

F = 50000 N

Area = 1000 cm^2

Covert it in m^2

= 0.1 m^2

Using Formula

= 500000

Ffhjiiftknv fsuvxfhkj

Where's the QUESTION it's only the INFO

Allowable BEARING PRESSURE: It is the MAXIMUM soil pressure WITHOUT any shear failure or settlement failure. ... These values don't consider importantfactors affecting the bearing capacitysuch as the shape, width, depth of footing, location of water table,strength and compressibility of the soil.
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Industrial waste is the waste produced by industrial activity which includes any MATERIAL that is rendered useless during a manufacturing process such as that of factories, industries, mills, and mining OPERATIONS. It has EXISTED SINCE the start of the Industrial Revolution.[1] Some examples of industrial wastes are chemical solvents, pigments, sludge, metals, ash, paints, sandpaper, paper products, industrial by-products, and radioactive wastes.

Toxic waste, chemical waste, industrial solid waste and municipal solid waste are DESIGNATIONS of industrial wastes. Sewage treatment plants can treat some industrial wastes, i.e. those consisting of conventional pollutants such as biochemical oxygen demand (BOD). Industrial wastes containing toxic pollutants require specialized treatment systems.

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Pollution control DEVICES are a series of devices that work to prevent a variety of different pollutants, both gaseous and solid, from entering the atmosphere primarily out of industrial smokestacks. These control devices can be separated into two broad categories - devices that control the amount of particulate MATTER escaping into the environment and devices that control acidicgas emissions. It is important to understand that the extraction methods for each specific type of pollutant can differ, so the only the major methods are discussed. Although complex, these devices have shown to be effective in the past with the overall levels of emissions for many pollutants dropping with the implementation of these control devices.

To look at data showing how the emission levels of a variety of pollutants has changed over time, click here.

Particulate Control

Specific machinery is used to remove particulate matter from flue gases. Much of this separation uses physical means of separation and not chemical separation techniques simply because particulate matter is large enough to be "caught" in this manner. Below are some of the basic ways that particulate matter can be extracted.

Electrostatic Precipitators

main article

An electrostatic precipitator is a type of filter that uses static electricity to remove sootand ash from exhaust fumes before they exit the smokestacks.[2] Unburned particles of carbon in SMOKE are pulled out of the smoke by using static electricity in the precipitators, leaving clean, hot air to escape the smokestacks.[2] It is vital to remove this unreacted carbon from the smoke, as it can damage buildings and harm human health - especially respiratory health.

CYCLONE Separators

main article

A cyclone separator is a separation device that uses the principle of inertia to remove particulate matter from flue gases.[3] In these separators, dirty flue gas enters a chamber containing a vortex, similar to a tornado. Because of the difference in inertia of gas particles and larger particulate matter, the gas particles move up the cylinder while larger particles hit the inside wall and drop down. This separates the particulate matter from the flue gas, leaving cleaned flue gas.

Fabric Filters

main article

Fabric filters are one fairly simple method that can be used to remove dust from flue gases. In some gases they can also remove acidic gases if they utilize basic compounds. This method simply uses some SORT of fabric - generally felt is used as a woven cloth would allow dust to make its way through - is placed so that flue gasses must pass through it before exiting the smokestacks.[4]When the gas passes through, dust particles are trapped in the cloth.

Gas Control

More intense chemical methods of separation are generally required to separate polluting gases from the flue gas. However, this extraction is important as many acidic gases in flue gas contribute to acid rain. Below are some of the basic ways that gases can be extracted.

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Swimming pool sanitation is the PROCESS of ensuring HEALTHY conditions in swimming pools, hot tubs, plunge pools, and similar recreational water venues. Proper sanitation is NEEDED to maintain the visual clarity of water and to prevent the TRANSMISSION of infectious WATERBORNE diseases.
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Standard for noise pollution

The Government has chalked out standard for noise pollution The ambient noise standards have been notified for different categories like, industrial, COMMERCIAL and residential areas and silence zones. 

It is 55 dB (A) Leq during ‘day time’ and 45 dB (A) Leq during ‘night time’ for residential areas. ‘Day time’ and ‘night time’ mean 6.00 am to 10.00 pm and 10.00 pm to 6.00 am, respectively. dB (A) Leq denotes the time weighted average of the level of sound in decibels on scale A which is relatable to human hearing. 

The Central Pollution Control Board (CPCB) and the State Pollution Control Boards (SPCBs)/ Pollution Control Committees (PCCs) have been carrying out sporadic noise MONITORING in urban areas, mainly during festivals such as Deepawali. A few SPCBs and PCCs have initiated regular noise monitoring since 2008-2009. As per available data, the prescribed noise norms for various areas are exceeded at many locations. However, a definite trend can not be ascertained since limited data is available. The CPCB has undertaken the task to establish a National Ambient Noise Monitoring Network. 

This information was given by the Minister of State for Environment and Forests (independent charge) Shri Jairam Ramesh in a written reply to a question by Shri N Balaganga in Rajya Sabha today. 

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Buck boost CONVERTER uses an INDUCTOR DRIVEN by a square wave. the inductor stores energy initially. A diode switch blocks CURRENT flow to the capacitor at first. the square wave is turned off and the inductor keeps current flowing to the capacitor which charges through the diode.
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The ENERGY is ALREADY PRESENT in our UNIVERSE there is only to FOUND that

1) the image formed is of the same SIZE as that of object.
2) the image is formed at the same distance BEHIND the mirror as the object is in FRONT of it.
3) the right APPEARS to be in left and left appears to be in right due to LATERAL inversion.
4) the image formed is erect

Landfill Fundamentals

Landfill is the discarding of waste into or onto land. Historically, landfills are the oldest and the most common method of waste treatment and disposal in many places AROUND the world. Today, landfill sites are engineered, operated, and monitored under strict technical standards to ensure compliance with federal regulations and to protect the environment from noxious waste.

Most types of WASTES may be discarded off through landfills; HOWEVER, due to environmental safety concerns, most municipalities have prohibited unregulated landfills and burning in a BID to prevent any linkage between the wastes and the SURROUNDING environment. Particularly the groundwater supplies, streams, and airways.

HEY BUDDY HERE IS YOUR ANSWER

Noise pollution affects both health and behavior. Unwanted sound (noise) can damage physiological health. Noise pollution can causehypertension, HIGH stress levels, TINNITUS, hearing LOSS, sleep disturbances, and other harmful effects.

HOPE IT'S HELPFUL FOR YOU

In ORDER to become black HOLE its radius should be very small and its DENSITY should be HIGH.

# ANSWER- 15×10^9 joules.

# Explanation-

To solve this kind of problems you must take CARE to put everything in same system of units preferably SI units.

# Given-

Pressure P = 3×10^5 N/sq.m

Pumped water dV = 50000cu.m = 5×10^4cu.m

# Solution-

WORK done for PUMPING water is givem by

W=PdV

W = 3×10^5 × 50000

W = 15×10^9 Joules.

# TOTAL work done in pumping the water out is 15×10^9 joules.

Hope that cleared it...

What is your QUESTION???

2/5MR^2

therefore, ANSWER is : R^2

A salt is any compound which can be derived from the neutralization of an acid and a base. The word "neutralization" is USED because the acid and base PROPERTIES of H+ and OH- are destroyed or NEUTRALIZED. ... A neutralization is a type of double REPLACEMENT

There are four DIFFERENT WAYS in which HAZARDOUS wastes can be finally disposed.Landfill disposal.Incineration.Dumping at sea.Underground disposal.

It will INCREASE by 4 TIMEA

Compelete your QUESTION

Hey mate here is your ans.
importance of MATHS is to SOLVE numericals be coz jub AAP HIGHER classes mein jaoge tab jaroorat paregi

Given: A piece of COPPER having an internal cavity weighs 264 g in air and 221 g in water. the density of copper is 8.8 g cm-3 what is the volume of the cavity ?

As 

loss in the WEIGHT of the piece in the water will refer to the buoyancy acting in the UPWARD DIRECTION, FB = V ρ g = 43 g

=>  V = 4.39 CM3

Volume of copper piece = 264/8.8 = 30 cm3

Hence,

Volume of the cavity = (30-4.39) cm3 = 25.61 cm3

The SOURCES of INDUSTRIAL WASTE WATER are

Hii
there is answer !!

I hope you help !!

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=========================================



When we walk on hard ground, the force with which our foot pushes the ground is equal to the force with which the ground pushes the foot. SINCE, the ground is hard it doesn’t change its shape and we walk EASILY. While in SAND, a part of the force by our foot is used up in displacing the sand. So, the reaction force now is smaller. So, it becomes difficult to walk on sand.

__

Suppose:

Mass of bullet = m

Mass of gun = M

Velocity of bullet = V

Recoil velocity of the gun = V

Initially the gun and bullet were at rest, so, the initial momentum of the system is, Pi = 0

After firing the momentum of the system = momentum of gun + momentum of bullet

=> Pf = MV + mv

By conservation of momentum,

Pf = Pi 

=> MV + mv = 0

=> V = -(mv)/M

The negative sign indicates that the direction of velocity of gun is opposite to that of the bullet.

For a lighter gun, M is smaller, so V will be large. And for a heavy gun M is larger and V will be smaller.

So, recoil velocity of a heavier gun is comparatively smaller.

___

We have,

Initial velocity,  u =3 m/s

Final velocity, v = 7 m/s

Mass of the BODY, m = 5 kg

Now, acceleration, a = (v - u)/t = (7 - 3)/2 = 2 m/s2

So applied force, F = ma = (5)(2) = 10 N

If the force was applied for 5 s the final velocity would be,

v = u + at

=> v = 3 + (2)(5) = 13 m/s

___

To get answers of all your questions, you may post them separately

A neutralization is a type of double REPLACEMENT reaction. A salt is the PRODUCT of an acid-base reaction and is a much BROADER TERM then common table salt as shown in the FIRST reaction. The following are some examples of neutralization reactions to form salts.

1.listed WASTE,2.universal waste

A sluice (from the Dutch "SLUIS") is a WATER channel controlled at its head by a gate. ...Sluice gates commonly control water levels and flow rates in rivers and canals. They are also used in WASTEWATER treatment plants and to recover minerals in mining operations, and in watermills.

A RELIEF valve or pressure relief valve is a type of safety valve used to control or limit the pressure in a system; pressure MIGHT otherwise build up and create a process upset, instrument or equipment failure, or fire. ... As the fluid is diverted, the pressure inside the vessel will stop rising

25 km = 25000 m
As the speed of car = 25000 m/hour
DISTANCE COVERED by the car in a MIN = 25000/60 = 416.67
Distance covered by the car in a sec = 6.95 m

We KNOW,
V=u+at
Squaring,
Or, V2=u2+2uat+a2t2
Or, v2=u2+2a(ut+1/2at2)
Or,v2=u2+2as(HENCE,s=ut+1/2at2)....