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Answered 23 hrs ago CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation

Let us assume that our galaxy consists of 2.5 x 1011 stars each of one solar mass. How long will a star... read more

Let us assume that our galaxy consists of 2.5 x 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky way to be 105 ly.

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Balakumar

Automobile Engineer

Answer:- Mass of our galaxy Milky Way, M = 2.5 × 1011 solar massSolar mass = Mass of Sun = 2.0 × 1036 kgMass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 × 1041 kgDiameter of Milky Way, d = 105 lyRadius of Milky Way, r = 5 × 104 ly1 ly = 9.46 × 1015... read more
Answer:-
 

Mass of our galaxy Milky Way, M = 2.5 × 1011 solar mass
Solar mass = Mass of Sun = 2.0 × 1036 kg
Mass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 × 1041 kg
Diameter of Milky Way, d = 105 ly
Radius of Milky Way, r = 5 × 104 ly
1 ly = 9.46 × 1015 m
r = 5 × 104 × 9.46 × 1015
= 4.73 ×1020 m
Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:
= ( 4π2r3 / GM)1/2
= [ (4 × 3.142 × 4.733 × 1060) / (6.67 × 10-11 × 5 × 1041) ]1/2
=   (39.48 × 105.82 × 1030 / 33.35 )1/2
= 1.12 × 1016 s
1 year = 365 × 324 × 60 × 60 s
1s = 1 / (365 × 324 × 60 × 60)  years
∴ 1.12 × 1016 s = 1.12 × 1016 / (365 × 24 × 60 × 60)  =  3.55 × 108 year

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Answered 20 hrs ago CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation

Does the escape speed of a body from the Earth depend on (a) the mass of the body, (b) the location from... read more

Does the escape speed of a body from the Earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?

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Gaurav

Tutor

Total energy of a moving body (of mass m)from earth's surface is given by the sum of potential and kinetic energy viz -(GMm/r)+0.5mv² To have enough vel so as to reach space, where the potential energy becomes negligible and velocity of body reaches 0, implies that if we use the above equation... read more

Total energy of a moving body (of mass m)from earth's surface is given by the sum of potential and kinetic energy viz   -(GMm/r)+0.5mv²

To have enough vel so as to reach space, where the potential energy becomes negligible and velocity of body reaches 0, implies that if we use the above equation and equate it to 0, we will get the escape vel.

Thus we see that escape vel should not be dependent on the mass of the body, but depends on vertical vel i.e direction of projextion, height from earth as well as location of earth(since earth is not completely spherical).

 

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Answered 1 day ago CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 6-System of Particles and Rotational Motion

Find the components along the x, y, z-axes of the angular momentum l of a particle, whose position vector... read more

Find the components along the x, y, z-axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z- component.

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Om Prakash Mishra

Tutor

Lx= YPz-ZPy, Ly =ZPx-XPz, Lz =XPy-YPx. If the particle X-Y plane then Z=0, and Pz=0. So only Z component is left. Lz=XPy-YPx
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Answered 23 hrs ago CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 6-System of Particles and Rotational Motion

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia... read more

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.

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Om Prakash Mishra

Tutor

(a) I = (2MR^2)/5 +MR^2 = 7MR^2/5 (Parallel axis theorem) (b) I = (MR^2)/4 + (MR^2)/4 =(MR^2)/2 (Perpendicular axis theorem.)
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Answered 1 day ago CBSE/Class 11/Science/Physics/Unit 2-Kinematics/Chapter 4-Motion in a Plane

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation... read more

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10 s apart is 30°, what is the speed of the aircraft? Time taken by aircraft from A to B is 10 s.

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Vaibhav Sharan

Tutor

Consider a point A vertical to the ground and plot a point B after 10 sec Speed = Dist/Time = AB/10 sec tan 30 = AB/3400 = 1/√3 => AB =3400/√3 m Speed = AB/10 = 340/√3 = 196.2990 m/sec
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Answered 23 hrs ago CBSE/Class 11/Science/Physics/Unit 3-Laws of Motion/Chapter 5-Laws of Motion

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms-1... read more

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms-1 what is the recoil speed of the gun?

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Balakumar

Automobile Engineer

Mass of the gun, M = 100 kg Mass of the shell, m = 0.020 kgMuzzle speed of the shell, v = 80 m/sRecoil speed of the gun = VBoth the gun and the shell are at rest initially.Initial momentum of the system = 0Final momentum of the system = mv – MVHere, the negative sign appears because the directions... read more

Mass of the gun, M = 100 kg

Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, v = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mv – MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
mv – MV = 0
∴ V = mv / M
= 0.020 × 80 / (100 × 1000) = 0.016 m/s

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Answered 1 day ago CBSE/Class 11/Science/Physics/Unit 2-Kinematics/Chapter 4-Motion in a Plane

State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed,... read more

State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

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Vaibhav Sharan

Tutor

Volume ScalarMass ScalarSpeed ScalarAcceleration VectorDensity ScalarNumber of Moles ScalarVelocity VectorAngular Frequency ScalarDisplacement VectorAngular... read more

Volume                      Scalar
Mass                          Scalar
Speed                        Scalar
Acceleration               Vector
Density                      Scalar
Number of Moles        Scalar
Velocity                     Vector
Angular Frequency      Scalar
Displacement             Vector
Angular Velocity         Vector

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Answered 1 day ago CBSE/Class 11/Science/Physics/Unit 2-Kinematics/Chapter 4-Motion in a Plane

Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear... read more

Pick out the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

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Vaibhav Sharan

Tutor

Work & Current
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Answered 1 day ago CBSE/Class 11/Science/Physics/Unit 2-Kinematics/Chapter 4-Motion in a Plane

Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power,... read more

Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

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Vaibhav Sharan

Tutor

Impulse
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Asked on 14 Sep CBSE/Class 11/Science/Physics/Unit 2-Kinematics/Chapter 4-Motion in a Plane

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ncert-solutions-class-11th-physics-chapter-4-motion-plane-5

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