Tags Physics

Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question

5.6k

Questions

Discussions

Active Users

Followers

New question posted

8 months ago

Physics Work, Energy and Power

6.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx²/2, where k is the force constant of the oscillator. For k = 0.5 N m^-1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

0 Follower 4 Views

New answer posted

8 months ago

Physics Work, Energy and Power

6.3 Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

0 Follower 194 Views

Payal Gupta

Contributor-Level 10

6.3 We know the total energy E is given by E = Kinetic energy (KE) + Potential energy (PE)

(a) In figure (a), we have at x=0, the potential energy is zero. So KE is positive. At x>a, the potential energy has a value greater than E, so the KE becomes 0. Thus the particle will not exist in the region x>a. Minimum total energy is zero.

(b) For the entire x-axis, PE >E, the KE of the object would be negative. Thus the particle will not exist in this region.

(c) In x=0 to x=a and x>b, PE is greater than E, so he KE has to be negative. The object cannot exist in this region.

(d) For x=-b/2 to x =-a/2 and x=a/2 to x=b/2, KE

...more

0 0

New answer posted

8 months ago

Physics Work, Energy and Power

6.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.

Compute the

(a) Work done by the applied force in 10 s

(b) Work done by friction in 10 s

(c) Work done by the net force on the body in 10 s

(d) Change in kinetic energy of the body in 10 s, and interpret your results

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

6.2 Given, mass of the body, m = 2 kg

Horizontal force applied, F = 7 N

Coefficient of friction, $?$ = 0.1

Acceleration, a = F/m = 7/2 = 3.5 m/s2

Frictional force, f = $? R = ? m g =$ 0.1 $* 2 * 9.807 =$ 1.96 N

Retardation produced by the frictional force, $a_{r}$ = -f/m = -1.96 /2 = 0.98 m/s2

The net acceleration by which the body moves forward

$a_{n}$ = a - $a_{r}$ = 3.5 – 0.98 = 2.52 m/s2

Distance moved by the body in 10 s is given by

s = ut + (1/2) $a_{n} t^{2}$ = 0 $* 10 + (\frac{1}{2}) * 2.52 * 10^{2}$ = 126 m

(a) Work done in 10 s is given by

W = Force $* d i s p l a c e m e n t = 7 * 126 = 882 J$

(b) Work done by friction in 10 s is given by

W = -f $* s = - 1.96 * 126 =$ - 247J

(c) Work done by the net force

...more

6.2 Given, mass of the body, m = 2 kg

Horizontal force applied, F = 7 N

Coefficient of friction, $?$ = 0.1

Acceleration, a = F/m = 7/2 = 3.5 m/s2

Frictional force, f = $? R = ? m g =$ 0.1 $* 2 * 9.807 =$ 1.96 N

Retardation produced by the frictional force, $a_{r}$ = -f/m = -1.96 /2 = 0.98 m/s2

The net acceleration by which the body moves forward

$a_{n}$ = a - $a_{r}$ = 3.5 – 0.98 = 2.52 m/s2

Distance moved by the body in 10 s is given by

s = ut + (1/2) $a_{n} t^{2}$ = 0 $* 10 + (\frac{1}{2}) * 2.52 * 10^{2}$ = 126 m

(a) Work done in 10 s is given by

W = Force $* d i s p l a c e m e n t = 7 * 126 = 882 J$

(b) Work done by friction in 10 s is given by

W = -f $* s = - 1.96 * 126 =$ - 247J

(c) Work done by the net force,

W = (F-f) $* s$ = (7 – 1.96 ) $* 126$ = 635 J

(d) Kinetic energy is given by the equation

KE = (1/2)m $v^{2}$ where v is the final velocity

From the equation v = u + at we get after 10 s, v = 0 + 2.52 $*$ 10 = 25.2 m/s

Final kinetic energy = (1/2) $* 2 * {25.2}^{2}$ = 635 J

Initial kinetic energy = (1/2) $* 2 * 0^{2}$ = 0

Change in kinetic energy = 635 – 0 = 635 J

Hence the work done by the net force = change in kinetic energy

less

0 0

New answer posted

8 months ago

Physics Work, Energy and Power

6.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket

(b) Work done by gravitational force in the above case

(c) Work done by friction on a body sliding down an inclined plane

(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity

(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest

0 Follower 86 Views

Payal Gupta

Contributor-Level 10

6.1 (a) While the person lifts a bucket out of a well by means of a rope tied to the bucket, the direction of both the force and the displacement are same, hence the work done is positive.

(b) While lifting the bucket, he works against gravity, but the work done by the gravitational force is downward, hence the work done is negative.

(c) The direction of motion of the object is in the opposite direction of the frictional force; hence the work done is negative.

(d) While a body moves on a rough horizontal plane, the frictional forces try to oppose the motion. But since the applied force maintains uniform velocity

...more

0 0

New answer posted

8 months ago

Physics Motion in a Plane

4.25 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 10.0 s apart is 30°, what is the speed of the aircraft?

0 Follower 317 Views

Vishal Baghel

Contributor-Level 10

4.25

Height of the aircraft from the ground = 3400m

Let A and B be the positions of the aircraft, making an angle of AOB = 30 $°$ . The perpendicular OC and OC is the height of the aircraft. Angles AOC = angle BOC = 15 $°$

In AOC, AC = OC $\tan$
$15 °$ = 3400 $\tan ? 15 °$ = 911.03m

AB = AC + CB = 2AC = 1822m

The distance of AB is covered in 10s, so the speed of the aircraft = 1822/10 m/s = 182.2 m/s

0 0

New answer posted

8 months ago

Physics Motion in a Plane

24 Read each statement below carefully and state, with reasons and examples, if it is true or false:

A scalar quantity is one that

(a) Is conserved in a process

(b) Can never take negative values

(c) Must be dimensionless

(d) Does not vary from one point to another in space

(e) Has the same value for observers with different orientations of axes

0 Follower 98 Views

Vishal Baghel

Contributor-Level 10

4.24

(a) Despite being a scalar quantity, energy is not conserved in elastic collisions. False

(b) Despite being a scalar quantity, the temperature can take negative values. False

(c) The total path length is a scalar quantity. Yet it has the dimension of length. False

(d) A scalar quantity such as gravitational potential can vary from one point to another point in space. False

(e) The value of a scalar does not vary for observers with different orientation of axis. True

0 0

New answer posted

8 months ago

Physics Motion in a Plane

4.23 For any arbitrary motion in space, which of the following relations are true:

(a) V average = (1/2) (v (t1) + v (t2))

(b) V average = [r (t2) - r (t1)] / (t2 – t1)

(c) V (t) = v (0) + a t

(d) R (t) = r (0) + v (0) t + (1/2) a t²

(e) A average = [v (t2) - v (t1)] / (t2 – t1)

(The 'average' stands for average of the quantity over the time interval t1 to t2)

0 Follower 97 Views

Vishal Baghel

Contributor-Level 10

4.23

(a) For any arbitrary motion of a particle average velocity cannot be expressed by this equation. False

(b) The arbitrary motion of the particle can be represented by this equation, True

(c) For arbitrary motion of the particle, the acceleration may also be non uniform. False

(d) The motion of the particle is arbitrary, acceleration of the particle may also be non-uniform, so can not represent the motion of the particle in space. False

(e) The arbitrary motion of the particle can be represented by the given equation. True

0 0

New answer posted

8 months ago

Physics Kinetic Theory

13.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

Substance	Atomic Mass (u)	Density ( $10^{3} K g m^{- 3})$
Carbon ( Diamond)	12.01	2.22
Gold	197.00	19.32
Nitrogen (liquid)	14.01	1.00
Lithium	6.94	0.53
Fluorine (liquid)	19.00	1.14

[Hint : Assume the atoms to be 'tightly packed' in a solid or liquid phase, and use the known value of Avogadro's number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

0 Follower

Payal Gupta

Contributor-Level 10

13.14 Let the atomic mass of a substance be = M and the density of the substance be = $?$

Avogadro's number, N = 6.023 $* 10^{23}$

Volume of N number of molecules = $\frac{4}{3} ? r^{3} N$ ……. (i)

Volume of one mole of a substance = $\frac{M}{?}$ …… (ii)

Equating (i) and (ii), we get

$\frac{4}{3} ? r^{3} N = \frac{M}{?}$

0 0

New answer posted

8 months ago

Physics Kinetic Theory

13.13 A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT]

where n₂, n₁ refer to number density at heights h₂ and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:

n₂ = n₁ exp [ -mg N_A ( $ρ$ - $ρ^{'}$ ) (h₂ –h₁₎/ ( $ρ$ RT)]

where $ρ$ is

...more

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT]

where n₂, n₁ refer to number density at heights h₂ and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:

n₂ = n₁ exp [ -mg N_A ( $ρ$ - $ρ^{'}$ ) (h₂ –h₁₎/ ( $ρ$ RT)]

where $ρ$ is the density of the suspended particle, and that of surrounding medium. [N_A is Avogadro's number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]

less

0 Follower 1 View

Payal Gupta

Contributor-Level 10

13.13 According to law of atmospheres, we have

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT] ….(i)

where $n_{1}$ is the number of density at height $h_{1}$ and $n_{2}$ is the number of density at height $h_{2}$

mg is the weight of the particle suspended in the gas column

Density of the medium = $?$

Density of the suspended particle = $?^{'}$

Mass of one suspended particle = m'

Mass of medium displaced = m

Volume of the suspended particle = V

According to Archimedes's principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:

Weight of the medium displaced – weight of the suspended particle = mg – m'g

= mg- V $?^{'} g$ = mg –

...more

13.13 According to law of atmospheres, we have

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT] ….(i)

where $n_{1}$ is the number of density at height $h_{1}$ and $n_{2}$ is the number of density at height $h_{2}$

mg is the weight of the particle suspended in the gas column

Density of the medium = $?$

Density of the suspended particle = $?^{'}$

Mass of one suspended particle = m'

Mass of medium displaced = m

Volume of the suspended particle = V

According to Archimedes's principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:

Weight of the medium displaced – weight of the suspended particle = mg – m'g

= mg- V $?^{'} g$ = mg – ( $\frac{m}{?}$ ) $?^{'} g$

= mg(1- $\frac{?^{'}}{?}$ ) ……(ii)

Gas constant R = $k_{B}$ N

$k_{B} = \frac{R}{N}$ …(iii)

Substituting from (ii) and (iii) in (i), we get

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT]

= n1 exp [ $[- m g (- \frac{?^{'}}{?}) (h 2 - h 1) \frac{N}{R T}]$ ]

= n1 exp[ $[- m g (? - ?^{'}) (h 2 - h 1) \frac{N}{R T ?}]$ ]

less

0 0

New answer posted

8 months ago

Physics Kinetic Theory

13.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^–1. Identify the gas.

[Hint : Use Graham's law of diffusion: R1/R2 = ( M₂ /M₁ )^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]

0 Follower 1 View

Payal Gupta

Contributor-Level 10

13.12 Rate of diffusion of hydrogen, $R_{1}$ = 28.7 cm³ s^–1

Rate of diffusion of another gas, $R_{2}$ = 7.2 cm³ s^–1

According to Graham's law of diffusion, we have:

$\frac{R_{1}}{R_{2}}$ = $\sqrt{\frac{M_{2}}{M_{1}}}$ , where $M_{1}$ = molecular mass of hydrogen = 2.02 g and $M_{2}$ is the molecular mass of the unknown gas

$M_{2} = M_{1} * ({\frac{R_{1}}{R_{2}})}^{2}$ = 2.02 $* ({\frac{28.7}{7.2})}^{2}$ = 32.09 = Molecular mass of Oxygen

Hence, the unknown gas is Oxygen.

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
684k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Physics

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience