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Thestarsare much much much farther away than any distance you canmoveon the Earth, so you shouldn't be able to see them "move" on theskyjust bymovingon the Earth. One thing to keep in mind, however, is that thestars do moveslowly over the course of the night.

stars will be near the horizon

near the horizon because you look through more atmosphere

Stars do not twinkle but appear to twinkle when seen from Earth. This is due to atmospheric refraction of light.Stars seem to twinkle because their light must pass through pockets of Earth's atmosphere that vary in temperature and density, and it's all very turbulent. On rough nights, a star appears to shift position constantly as itslightis refracted this way and that.

stars do not twinkle really,they just appear to twinklewhen seen from the surface of the area.the stars

stars don't twinkle,they just appear to twinklewhen seen from the surface of the earth.the stars twinklein the night sky because of the effects of our atmosphere.when star light enters our atmosphere.it is effected by winds in the atmosphere and by areas with different temperatures and densities.

Moon has no atmosphere and weather because the value of acceleration due to gravity ‘g’ on surface of moon is small. Therefore, the value of escape velocity on the surface of the moon is small . The molecules of the atmospheric gases on the surface of the moon have thermal velocities greater than the escape velocity. That is why all the molecules of gases have escaped and there is no atmosphere on moon.

Ans:(b) contact is made

It is defined as the minimum distance between the eye lens and the object to form a clear image.Least distance of distinct vision for a normal human being is 25cm.

a. Deala with conservation of Charge

the unit of electric flux density is C m^-2

static electric fields are conservative in nature.E=-dV/dr

A conservative force is a force with the property that the total work done in moving a particle between two points is independent of the taken path. ... When an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken.

As you say whenpushing a book, the work that you do "against friction" is apparently lost - it is certainly not available to the book as kinetic energy! Forces that do not store energy are called nonconservative or dissipative forces. Friction is a nonconservative force, and there are others. Any friction-type force, like air resistance, is a nonconservative force. The energy that it removes from the system is no longer available to the system for kinetic energy.

there is rotation also involved

answer is 1.73 m/s

Infinity!In a freely orbiting satellite which is revolving around the earth, the net value of acceleration due to gravity is zero and hence we experience zero gravity inside the spacecraft.

And, the time period of a simple pendulum is is inversely proportional to the square root of acceleration due to gravity. Time period=2 ×pi× (l/g)^1/2 (where l is length of the simple pendulum and g is acceleration due to gravity)

The value of g is zero inside a spacecraft, and in the equation g is in the denominator. And any number divided by zero is NOT DEFINED.

As per me, either the time period is not defined or infinity.

And since the time period cannot be not defined the answer is infinity which also implies that the pendulum won't oscillate.

Area of parallelogram is cross products of sides

= |A×B| = |(i+4j)×(3i+4j)| = |(4-12)k|= 8k

and modulus is = 8 units

option 2

its not 3i+4j its 3i ×4j

Then first on resolving 3i×4j = 12k

so area will be | (i+4j)×(12k) |= = 48i +12j

and modulus = |√48²+12²| = 49.47 units

option D is the ✓

so, there can be a printing error.. it should be 3i+4j

eliminate t from y , by putting t = x/5

we get y = (x/5)³+(x/5)²+(x/5) = x³+5x²+25x

b. MR^2/2

Answer:

Force required is 2kqa/r

Explanation:

Point kept at centre = q

Radius = R

Charge at the segment rdθ =ardθ

The horizontal component of force will be cancelled out as the ring will be symmetric about y axis

Vertical force =Fcosθ =kqardθ/r2Cosθ

=kqa/r cosθ dθ

Thus, total force = limit 0 to 90, 2∫ kqa/r cosθ dθ

= limit 0 to 90, 2kqa/r∫ cosθ dθ

= 2kqa/r

centre of mass =2

he will see that rain is falling with a speed of 50m/s making angle 60° with horizontal

please show with answer

is it correct?

so ur ans is 50m/s and it's 60° with horizontal

Therefore, the plane must travel relative to the air, west of North.

the voltmeter resistance is in parallel to 60Ω , that is 40Ω , so they add up to give req = 60*40/(60+40) = 24Ω

now , total resistance is 40+24 = 64Ω

so, current in the circuit is 6/64 = 3/32A

now voltage around voltmeter is 3/32*(24) = 9/4 = 2.25V

option 1

a is the answer .......

the voltage drop El = -ldi/dt = -5*(10^-3)*(10³) =-5v so, the voltage drop Vb= Va -5+15-5 => Vb-Va = 5V

2) is the correct answer

Let the current flowing in the resistance be i

so, Potential difference is 60-0 = 60v ( because at G wire is grounded hence potential is 0)

now total resistance = 64+32 = 96

so, i = v/R = 60/96 = 5/8 A

potential at J = iR1 = 5/8*(64) = 40v

so, difference is = 40-0 = 40V

option 1

the answer could be option d

Option (b) are correct

Torque about a point = Total force × perpendicular distance from the point to that force.

Let anticlockwise torque = + veAnd clockwise acting torque = –veForce acting at the point B is 15 N

Therefore torque at O due to this force= 15 × 6 × 10–2× sin 37°= 15 × 6 × 10–2× 3/5 = 0.54 N-m (anticlock wise)

Force acting at the point C is 10 NTherefore, torque at O due to this force= 10 × 4 × 10–2= 0.4 N-m (clockwise)

Force acting at the point A is 20 NTherefore, Torque at O due to this force = 20 × 4 × 10–2× sin30°= 20 × 4 × 10–2× 1/2 = 0.4 N-m (anticlockwise)Therefore resultant torque acting at ‘O’ = 0.54 – 0.4 + 0.4 = 0.54 N-m.

Answer;b)Explanation:

From the principle of conservation energy and angular momentumβ particle spin is 1/2 (= h/2π), so the change must be of the same order, however it is found that spin is either unchanged or it faces an integral change. Neutrino Hypothesis successfully explains the phenomena. β−is used for an e- andβ+is used for an e+. An excited proton rich nucleus gets stability by capturing K-shell orbital e-.

For aβ−decay, a neutron transforms into a proton within the nucleus as shown.

and for aβ+decay, aprotontransforms into a neutron within the nucleus.

Ans :- Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation, such as an alpha particle, beta particle with neutrino or only a neutrino in the case of electron capture, or a gamma ray or electron in the case of internal conversion.

Fossil fuels are the dead remain of plants and animals buried in the earth for millions of years. fossil fuel have high calorific value. easily availabletransportation is easylarge amount of electricity can be generated

thanks

As there is one octave so R bar Q+ sthanks