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the centripetal acceleration will become half

I have a doubt.H=(V^2/R)×tFrom the above formula, H=R^-1So, heating effect should be doubled.

a law stating that when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage.

Weight is given by the expression :

Weight=W=mg

so when g= acceleration due to gravity becomes zero then weight will be zero.

Conditions when weight will become zero:

1. Under a free fall

2. At the centre of the earth.[g=0]

3. In space, where zero gravity is there.

Standardized units of measure are important because if everyone was using different measurement systems, dimensions of different objects and inventions wouldn't be converted currently from them when the inventor to other inventors or manufacturers.

In 3 metre 3 is called magnitude and metre is called as unit.

The answer is to become worse

A fossil fuel is a fuel formed by natural processes, such as anaerobic decomposition of buried dead organisms, containing energy originating in ancient photosynthesis.

The de Broglie wavelength is the wavelength, λ, associated with a massive particle and is related to its momentum, p, through the Planck constant, hlamda=h/mv

volume of cube = (1.2x 10²)³ cm³ = (1.2)³ x 10^6 = 1.728 x 10^6 cm³

hence, option (a) is correct

volume of cube = a³ (1.2)³= 1.728m³1 m³ = 10^6 cm³volume = 1.728 × 10^6 cm³

At NTP temp = 293K and pressure is 1 atm, y= 1.4 and density is 1.29kg/m3

also P/density = RT and the value of R = 286.9J/k-kg

speed of sound in air is given by The speed of sound in air at20oC(293.15 K)and absolute pressure1 barcan be calculated as

c= (1.4 (286.9 J/K kg) (293.15 K))1/2

=343.1(m/s)

d. 17m

the wavelengthrange for hearing is 0.0172 m to 17.2 m.

thanque

answer 1 option is right

Let the charge on each object is q Given, separation between them , r = 5cm = 0.05 m force act between them , F = 0.144N [ repulsive force ]

Now, according to Coulombs law, F = Kq²/r² 0.144 = 9 × 10⁹ × q²/(0.05)² 0.016 × (0.05)² × 10⁻⁹ = q² Taking square root both sides, 0.4 × 0.05 × 10⁻⁵ = q q = 0.02 × 10⁻⁵ C = 0.2 × 10⁻⁶ C

Hence, magnitude of each charge = 0.2μC

the reflecting object should be V/20 m from the source of sound

It can be difficult to distinguish age-related hearing loss from hearing loss that can occur for other reasons, such as long-term exposure to noise. Noise-induced hearing loss is caused by long-term exposure to sounds that are either tooloudor last too long.

thank you

Sound is a form of energy which produces a sensation of hearing in our ears.

The speed of sound in air at 0℃ = 332m/s

If it increases at the rate of 0.6m/s per degree,

Speed at x℃ = (332 + 0.6x)℃

Thus 344 = 332 + 0.6x344 - 332 = 0.6x12 = 0.6xx = 12/ 0.6 = 20℃

Temperature is 20℃

Consider a satellite of mass m revolving round the Earth at a, height 'h' above the surface of the Earth.

Let M be the mass and R be the radius of the Earth.

The satellite is moving with velocity V and the radius of the circular orbit isr=R+hr=R+h. Centripetal force = Gravitational force

∴Mv2cr=GMmr2∴Mvc2r=GMmr2

∴v2c=GMr∴vc2=GMr

∴vc=√GMR+h∴vc=GMR+h

This is the expression for critical velocity of a satellite moving in a circular orbit around the Earth,

We know that,

gh=GM(R+h)2gh=GM(R+h)2

GM=gh(R+h)2GM=gh(R+h)2

Substituting in equation (1), we get

∴vc=√gh(R+h)2R+h∴vc=gh(R+h)2R+h

∴vc=√gh(R+h)∴vc=gh(R+h)

whereghghis the acceleration due to gravity at a height above the surface of the Earth.

The formula relating refractive index (μ)and critical angle (C) is given by :μ=1/sinCμ=1/sin45 =√2

1

2

{1/(10+10)} + { 1/C } = 1/15 C= 60

1/20+1/c=1/151/c=1/15-1/201/c=1/60c=60

c= 60 is the correct ans

60 is the correct answer

c= 60 is the right answer

60 is the right answer of the following

wrong

Critical Velocity (Vc)

l The minimum velocity required to revolve in a circular orbit around a planet is called critical velocity.

Vc=√[ ( G M ) / R ]

_ _ _ _ _ ( 1 )Where ,

G = Gravitational Constant

M = Mass of planet

R = Radius of planet

The density ( ρ ) of a body can be define as mass enclosed per unit volume

Therefore,

ρ = M / V

i.e.

M = ρ V

Volume of spherical planet is

V = ( 4 / 3 ) π R^3

The mass of planet can be given as

M = ρ ( 4 / 3 ) π R^3

Putting this value in equation ( 1 ) we get,

Vc=√[ ( ( 4 / 3 ) G ρ π R^3) / R ]

Vc= 2√[ ( G ρ π R^2) / 3 ]

Vc= 2R√[( Gρ π ) / 3 ]

wrong hai answer

Critical Velocity (Vc)

The minimum velocity required to revolve in a circular orbit around a planet is called critical velocity.

Vc=√[ ( G M ) / R ]

_ _ _ _ _ ( 1 )

Where ,

G = Gravitational Constant

M = Mass of planet

R = Radius of planet

The density ( ρ ) of a body can be define as mass enclosed per unit volume

Therefore,

ρ = M / V

i.e.

M = ρ V

Volume of spherical planet is

V = ( 4 / 3 ) π R3

The mass of planet can be given as

M = ρ ( 4 / 3 ) π R3

Putting this value in equation ( 1 ) we get,

Vc=√[ ( ( 4 / 3 ) G ρ π R3) / R ]

Vc= 2√[ ( G ρ π R2) / 3 ]

Vc= 2R√[( Gρ π ) / 3 ]

As the sensation of sound persists in our brain for about 0.1 s, to hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1s. Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least (344 m/s) ×0.1 s = 34.4 m. Thus, for hearing distinct echoes, the minimum distance of the obstacle from the source of sound must be half of this distance. I.e., 34.4/2 = 17.2 m

the angle of incidence beyond which rays of light passing through a denser medium to the surface of a less dense medium are no longer refracted but totally reflected.

How do a critical angle and a refractive index relate?

Originally Answered: What is the relation between refractive index and critical angle?The critical angle is that angle at which, if a ray is incident while passing from a denser to a rarer medium, the angle of refraction is 90o.

⇒sinicrsin90o=1μ where,

sinicr is the critical angle, and,

μ is the refractive index for a ray passing from the rarer to the denser medium.

⇒sinicr=1μ

⇒icr=arcsin(1μ).

Let us take an example of light passing from water to air. The refractive index for light passing from air to water is 43.

⇒icr=arcsin(34)=48.59o.

critical wavelength: The free-spacewavelengththat corresponds to thecriticalfrequency. Note: Thecritical wavelengthis equal, in meters, to the speed of light (3 × 108m/s) divided by thecriticalfrequency in hertz

critical wavelength:The free-spacewavelengththat corresponds to thecritical frequency.Note:The critical wavelength is equal, in meters, to thespeed of light(3 × 108m/s) divided by the critical frequency inhertz.

This HTML version of FS-1037C was last generated on Fri Aug 23 00:22:38 MDT 1996

90% of the area under the absorbance curve

critical wavelength:The free-spacewavelengththat corresponds to thecritical frequency.Note:The critical wavelength is equal, in meters, to thespeed of light(3 × 108m/s) divided by the critical frequency inhertz.

The wavelength at which the summed absorbance reaches 90% of total absorbance is defined as the 'critical wavelength' and is considered to be a measure of the breadth of sunscreen protection. Filters are then classified as 'broad spectrum', having a significant part of their absorbance in the UVA, when the critical wavelength is longer than 370 nm.

Thewavelengthat which the summed absorbance reaches 90% of total absorbance is defined as the 'critical wavelength' and is considered to be a measure of the breadth of sunscreen protection.

In seriesequivalent will be= 12*12*12/12+12+12= 48uf

in parallel= 12+12+12= 36 uf

Diamond (Because diamonds have a high index of refraction (about 2.3), the critical angle for the total internal reflection is only about 25 degrees. Incident light therefore strikes many of the internal surfaces before it strikes one less than25 degrees and then emerges.)

As the sensation of sound persists in our brain for about 0.1 s, to hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1s. If we take the speed of sound to be 344 m/s at a given temperature, say at 22 ºC in air, sound must go to the obstacle and reach back the ear of the listener on reflection after 0.1s.Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least (344 m/s) ×0.1 s = 34.4 m. Thus, for hearing distinct echoes, the minimum distance of the obstacle from the source of sound must be half of this distance. I.e., 34.4/2 = 17.2mhence 17 metres and more

For hearing distinct echoes,the minimum distance of the obstacle from the source of sound must be half of this distance that is 17.2m.