By QB365 on 31 Dec, 2022
QB365 provides a detailed and simple solution for every Possible Questions in Class 11 Physics Subject - Important 3 Mark English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
11th Standard
Physics
Answer All The Questions
In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 sec. If the speed of sound in water is 1460 ms-1. What is the distance of enemy submarine?
If the value of universal gravitational constant in SI is 6.6 x 10−11 Nm2 kg−2, then find its value in CGS System?
Calculate the average velocity of the particle whose position vector changes from \(\overrightarrow { { r }_{ 1 } } =5\hat { i } +6\hat { j } \) to \(\overrightarrow { { r }_{ 2 } } =2\hat { i } +3\hat { j } \) in a time 5 second.
The following velocity-time graph represents a particle moving in the positive x-direction. Analyse its motion from 0 to 7 s. Calculate the displacement covered and distance travelled by the particle from 0 to 2 s.
Given two vectors \(\vec A=2\hat i+4\hat j+5\hat k\) and \(\vec B=\hat i+3\hat j+6\hat k\). Find the product \(\vec A.\vec B\), and the magnitudes of \(\vec A\) and \(\vec B\) . What is the angle between them?
A particle moves in a circle of radius 10 m. Its linear speed is given by v = 3t where t is in second and v is in ms-1.
(a) Find the centripetal and tangential acceleration at t = 2 s.
(b) Calculate the angle between the resultant acceleration and the radius vector.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob into components. What is the acceleration experience by the bob at an angle ፀ.
A block of mass m slides down the plane inclined at an angle 60° with an acceleration \(\frac { g }{ 2 } \). Find the coefficient of kinetic friction.
An object of mass 1 kg is falling from the height h = 10m. Calculate
(a) The total energy of an object at h = 10 m.
(b) Potential energy ofthe object when it is at h = 4 m.
(c) Kinetic energy of the object when it is at h = 4 m.
(d) What will be the speed of the object when it hits the ground?
(Assume g = 10 m s-2)
From a uniform disc of radius R, a small disc of radius \(\frac{R}{2}\) is cut and removed as shown in the diagram. Find the center of mass of the remaining portion of the disc.
A solid sphere is undergoing pure rolling. What is the ratio of its translational kinetic energy to rotational kinetic energy?
Explain the variation of g with depth from the Earth’s surface.
Two particles of masses m1 and m2 are placed along the x and y axes respectively at a distance ‘a’ from the origin. Calculate the gravitational field at a point P shown in figure below.
A solid sphere has a radius of 1.5 cm and a mass of 0.038 kg. Calculate the specific gravity or relative density of the sphere.
When you mix a tumbler of hot water with one bucket of normal water, what will be the direction of heat flow? Justify.
During a cyclic process, a heat engine absorbs 500 J of heat from a hot reservoir, does work and ejects an amount of heat 300 J into the surroundings (cold reservoir). Calculate the efficiency of the heat engine?
Why does heat flow from a hot object to a cold object?
Consider a particle undergoing simple harmonic motion. The velocity of the particle at position x1 is v1 and velocity of the particle at position x2 is v2. Show that the ratio of time period and amplitude is
\(\frac { T }{ A } =2\pi \sqrt { \frac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 }x }_{ 2 }^{ 2 }-{ { v }_{ 2 }^{ 2 }x }_{ 1 }^{ 2 } } } \)
Let f be the fundamental frequency of the string. If the string is divided into three segments l1, l2 and l3 such that the fundamental frequencies of each segments be f1, f2 and f3, respectively. Show that \(\frac { 1 }{ f } =\frac { 1 }{ { f }_{ 1 } } +\frac { 1 }{ { f }_{ 2 } } +\frac { 1 }{ { f }_{ 3 } } \)
Define Dimensional Constant. Give example.
How many parallactic second are there in one Astronomical unit?
Given data:
1 parallactic second = 3.08\(\times\)1016m
1 Astronomical unit = 1.496\(\times\)1011m
"Technological advancements owe a great deal to the developments in physics" - Explain.
A body under the action of a force \(\vec { F } =(4\vec { i } -8\vec { j } +8\vec { k } )\)N acquires an acceleration of 2 ms-2. Find the mass of the body
What are non-inertial frames of reference?
When a force \(\overrightarrow{\mathrm{F}}=(8 \overline{\mathrm{i}}-10 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}) \mathrm{N}\) produces an acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) on a body. Calculate the mass of the body.
A body is displaced 10 \(\hat{j}\) under the force of \(-2\hat{j}+15\hat{j}+6\hat{i}\ N.\) Calculate the work done.
Can a body have energy without momentum?
A railway carriage of mass 9000 kg moving with a speed of 36 kmph collides with a stationary carriage of the same mass. After the collision, the carriages get coupled and move together. What is their common speed after the collision? What type of collision is this?
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, how much electric power is consumed by the pump. The efficiency of the pump is 30%.
State the conditions so that transfer of kinetic energy during a collision.
Answers
The speed of sound in water v = 1460 ms-1
Time taken t = 80 s
\(=\frac{2 d}{t} \)
2d = u x t
= 1460 x 80
2d =11680 m
\(\therefore \text { Distance of enemy submarine } d =\frac{11680}{2} \)
= 58.40 km
Let GSI be the gravitational constant in the SI system and Gcgs in the cgs system. Then
GSI = 6.6 ×10-11 Nm2 kg−2
Gcgs = ?
\( n_{2} =n_{1}\left[\frac{M_{1}}{M_{2}}\right]^{a}\left[\frac{L_{1}}{L_{2}}\right]^{b}\left[\frac{T_{1}}{T_{2}}\right]^{c} \)
\(\mathrm{G}_{c g s} =\mathrm{G}_{\mathrm{SI}}\left[\frac{M_{1}}{M_{2}}\right]^{a}\left[\frac{L_{1}}{L_{2}}\right]^{b}\left[\frac{T_{1}}{T_{2}}\right]^{c}\)
M1 = 1 kg L1 = 1m T1 = 1s
M2 = 1 g L2 = 1 cm T2 = 1s
The dimensional formula for G is M-1L3T-2
a = -1 b = 3 and c = -2
\( G_{c g s}=6.6 \times 10^{-11}\left[\frac{1 k g}{1 g}\right]^{-1}\left[\frac{1 m}{1 \mathrm{~cm}}\right]^{3}\left[\frac{1 s}{1 s}\right]^{-2} \)
\(=6.6 \times 10^{-11}\left[\frac{1 \mathrm{~kg}}{10^{-3} \mathrm{~kg}}\right]^{-1}\left[\frac{1 \mathrm{~m}}{10^{-2} \mathrm{~m}}\right]^{3}\left[\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right]^{-2}\)
= 6.6 × 10−11 × 10−3 x 106 × 1
Gcgs = 6.6 × 10-8 dyne cm2 g-2
\(\overrightarrow { { v }_{ 1 } } =5\hat { i } +6\hat { j } \)
\(\overrightarrow { { v }_{ 2 } } =2\hat { i } +3\hat { j } \)
\(\therefore \Delta \vec{r} =\vec{r}_{2}-\vec{r}_{1}=2 \hat{i}+3 \hat{j}-5 \hat{i}-6 \hat{j}
\)
\(=-3 \hat{i}-3 \hat{j}
\)
\(\Delta t =5 \mathrm{sec}
\)
\(\therefore \Delta v_{\text {avg }} s =\frac{\Delta \vec{r}}{\Delta t}
\)
\(=\frac{-3}{5}(\hat{i}+\hat{j})\)
It is a uniform motion
Distance travelled from 0 to 2 s= Area of the shaded portion
\(=\frac{1}{2} b_{1} h_{1}+\frac{1}{2} b_{2} h_{2}=\frac{1}{2} \times 1.5 \times 2+\frac{1}{2} \times 0.5 \times 1 \)
= 1.50 + 0.25 = 1.75m
Displacement
\(=-\frac{1}{2} b_{1} h_{1}+\frac{1}{2} b_{1} h_{2}
\)
\(=-\frac{1}{2} \times 1.5 \times 2+\frac{1}{2} \times 0.50 \times 1 \)
= -1.50 + 0.25 = -1.25 m
0-1 s The velocity increases in the negative direction
1 s-2 s Velocity increases in the positive direction. Change of velocity is 1-(-2)=3 ms-1
2 s-5 s Velocity does not change. So there is no acceleration.
5 s-6 s Velocity reduces to zero.
6 s-7 s The body is at rest.
\(\vec A.\vec B\) = 2 + 12 + 30 = 44
Magnitude A = \(\sqrt{4+16+25}=\sqrt{45}\) units
Magnitude B = \(\sqrt{1+9+36}=\sqrt{46}\) units
The angle between the two vectors is given by
\(\theta=\cos^{-1}(\frac{\vec A.\vec B}{AB})=\cos^{-1}(\frac{44}{\sqrt{45}\times \sqrt{46}})=\cos^{-1}(\frac{44}{45.49})=\cos^{-1}(0.967)\)
\(\therefore \theta \cong { 15 }^{ o }\)
The linear speed at t = 2 s
v = 3t = 6 ms-1
The centripetal acceleration at t = 2 s is
\(a_c=\frac{v^2}{r}=\frac{(6)^2}{10}=3.6\ ms^{-2}\)
The tangential acceleration is \(a_t=\frac{dv}{dt}=3\ ms^{-2}\)
The angle between the radius vector with resultant acceleration is given by
\(tan\theta=\frac{a_t}{a_c}=\frac{3}{3.6}\)= 0.833
\(\theta=tan^{-1}(0.833)\) = 0.69 radian
In terms of degree \(\theta=0.69\times57.17^0\approx40^0\)
(i) Tangential acceleration = g sin ፀ
(ii) mg cos ፀ acts along OP (outwards)
(iii) tension (T) acts along PO (inwards)
Net force on the body at P acting along
PO = T- mg cos ፀ
This must provide the necessary centripetal force. \(\frac{mv^2}{r}\)
\(\therefore T -mg \ cos \ \theta=\frac{mv^2}{r} \)
Centripetal acceleration (a⊥) = \(\frac { T-mg \ cos\theta }{ m } \)
Kinetic friction comes to playas the block is moving on the surface.
The forces acting on the mass are the normal force perpendicular to surface, downward gravitational force and kinetic friction fk along the surface.
mg sinθ - fk = ma
But a = g/2
mg sin 600 - fk = mg/2
\(\frac { \sqrt { 3 } }{ 2 } \)mg - fk = mg/2
fk = mg\(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } \right) \)
fk =\(\left( \frac { \sqrt { 3 } -1 }{ 2 } \right) mg\)
There is no motion along the y-direction as normal force is exactly balanced by the mg cos θ
mg cosθ = N = mg/2
fk = μkN = μk mg/2
μk = \(\frac { \left( \frac { \sqrt { 3 } -1 }{ 2 } \right) mg }{ \frac { mg }{ 2 } } \)
uk = \(\sqrt { 3 } \)-1
(a) The gravitational force is a conservative force. So the total energy remains constant throughout the motion. At h = 10 m, the total energy E is entirely potential energy.
E = U = mgh = 1 × 10 × 10 = 100 J
(b) The potential energy of the object at h = 4 m is
U = -mgh = 1 \(\times\) 10 \(\times\) 4 = 40 J
(c) Since the total energy is constant throughout the motion, the kinetic energy at h = 4m must be KE = E - U = 100 - 40 = 60 J.
Alternatively, the kinetic energy could also be found from velocity of the object at 4 m. At the height 4 m, the object has fallen through a height of 6 m.
The velocity after falling 6 m is calculated from the equation of motion,
\(v=\sqrt { 2gh } =\sqrt { 2\times 10\times 6 } =\sqrt { 120 } \) ms-1
v2 = 120
The kinetic energy is KE = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } \times 1\times 120=60J\)
(d) When the object is just about to hit the ground, the total energy is completely kinetic and the potential energy, U = 0.
\(E=KE=\frac { 1 }{ 2 } { mv }^{ 2 }=100\quad J\)
\(v=\sqrt { \frac { 2 }{ m } KE } =\sqrt { \frac { 2 }{ 1 } \times 100 } =\sqrt { 200 } =14.12\) ms-1
Let us consider the mass of the uncut full disc be M. Its center of mass would be at the geometric center of the disc on which the origin coincides.
Let the mass of the small disc cut and removed be m and its center of mass is at a position \(\frac{R}{2}\) to the right of the origin as shown in the figure.
Hence, the remaining portion of the disc should have its center of mass to the left of the origin say at a distance x. We can write from the principle of moments,
\((M-m)x=(m)\frac{R}{2}\)
\(x=(\frac{m}{M-m})\frac{R}{2}\)
If \(\sigma\) is the surface mass density (i.e. mass per unit surface area.), \(\sigma=\frac{M}{\pi R^{2}}\); then, the mass m of small disc is
m = surface mass density\(\times\) surface area
\(m=\sigma \times \pi (\frac{R}{2})^{2}\)
\(m=(\frac{M}{\pi R^{2}})\pi (\frac{R}{2})^{2}=\frac{M}{\pi R^{2}}\pi \frac{R^{2}}{4}=\frac{M}{4}\)
substituting m in the expression for x
\(x=\frac{\frac{M}{4}}{(M-\frac{M}{4})}\times \frac{R}{2}=\frac{\frac{M}{4}}{(\frac{3M}{4})}\times \frac{R}{2}\)
\(x=\frac{R}{6}\)
The center of mass of the remaining portion is at a distance \(\frac{R}{6}\) to the left from the center of the disc.
The expression for total kinetic energy in pure rolling is,
KE = KETRANS + KEROT
For any object the total kinetic energy as per given equation
\(KE=\frac{1}{2}Mv^{2}_{CM}+\frac{1}{2}Mv^{2}_{CM}(\frac{K^{2}}{R^{2}})\)
\(KE=\frac{1}{2}Mv^{2}_{CM}(1+\frac{K^{2}}{R^{2}})\)
Then, \(\frac{1}{2}Mv^{2}_{CM}(1+\frac{K^{2}}{R^{2}})\)=\(\frac{1}{2}Mv^{2}_{CM}+\frac{1}{2}Mv^{2}_{CM}(\frac{K^{2}}{R^{2}})\)
The above equation suggests that in pure rolling the ratio of total kinetic energy, translational kinetic energy and rotational kinetic energy is given as,
KE:KETRANS KEROT::\((1+\frac{K^{2}}{R^{2}}):1:(\frac{K^{2}}{R^{2}})\)
Now, KETRANS: KEROT::1:\((\frac{K^{2}}{R^{2}})\)
For a solid sphere, \(\frac{K^{2}}{R^{2}}=\frac{2}{5}\)
Then, KETRANS: KEROT::1:\(\frac{2}{5}\) or KETRANS: KEROT:: 5:2
Variation of g with depth:
Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines -in Neyveli). Assume the depth of the mine as d. To calculate g' at a depth d, consider the following points.
The part of the Earth which is above the radius (Re - d) do not contribute to the acceleration. The e· result is proved earlier and is given as
g' = \(\frac { GM' }{ ({ R }_{ e }-d)^{ 2 } } \)
Here M' is the mass of the Earth of radius (Re - d)
Assuming the density of Earth p to be constant
\(\rho =\frac { M' }{ V' } \)
where M is the mass of the Earth and V its volume, Thus
\(\rho =\frac { M' }{ V' } \)
\(\frac { M' }{ V' } =\frac { M }{ V } \) and M' = \(\frac { M }{ V } V'\)
M' = \(\left( \frac { M }{ \frac { 4 }{ 3 } \pi { R }_{ e }^{ 3 } } \right) \left( \frac { 4 }{ 3 } \pi ({ R }_{ e }-d)^{ 3 } \right) \)
M'=\(\frac { M }{ { R }_{ e }^{ 3 } } \)(Re - d)3
g' = G\(\frac { M }{ { R }_{ e }^{ 3 } } \)(Re - d)3.\(\frac { 1 }{ ({ R }_{ e }-d)^{ 2 } } \)
g' = GM \(\frac { R_{ e }\left( 1-\frac { d }{ { R }_{ e } } \right) }{ { R }_{ e }^{ 3 } } \)
g' = GM \(\frac { \left( 1-\frac { d }{ { R }_{ e } } \right) }{ { R }_{ e }^{ 2 } } \)
Thus
g' = g \(\left( 1-\frac { d }{ { R }_{ e } } \right) \)
Here also g' < g. As depth increases, g' decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.
Gravitational field due to m1 at a point P is given by,
\(\overrightarrow { { E }_{ 1 } } =-\frac { Gm_{ 1 } }{ { a }^{ 2 } } \hat { j } \)
Gravitational field due to m2 at the point p is given by,
\(\overrightarrow { { E }_{ 2 } } =-\frac { Gm_{ 2 } }{ { a }^{ 2 } } \hat { i } \)
\({ \vec { E } }_{ total }=\frac { { Gm }_{ 1 } }{ { a }^{ 2 } } \hat { j } -\frac { { Gm }_{ 2 } }{ { a }^{ 2 } } \hat { i } \)
\(=-\frac { G }{ { a }^{ 2 } } \left( { m }_{ 1 }\hat { j } +{ m }_{ 2 }\hat { i } \right) \)
The direction of the total gravitational field is determined by the relative value of m1 and m2
When m1 = m2 = m
\({ \overrightarrow { E } }_{ total }=-\frac { Gm }{ { a }^{ 2 } } \left( \hat { i } +\hat { j } \right) \)
(\(\hat { i } +\hat { j } =\hat { j } +\hat { i } \) as vectors obeys commutation law).
\({ \overrightarrow { E } }_{ total }\)Points towards the origin of the co-ordinate system and the magnitude of \({ \vec { E } }_{ total }is\frac { { Gm } }{ { a }^{ 2 } } .\)
Radius of the sphere R = 1.5 cm
mass m = 0.038 kg
Volume of the sphere V = \(\frac{4}{3}\pi{R^2}\)
= \(\frac{4}{3}\)\(\times\)(3.14)\(\times\)(1.5\(\times\)10-2)3 = 1.413\(\times\)10-5m3
Therefore, density
\(\rho =\frac { m }{ V } =\frac { 0.038kg }{ 1.413\times { 10 }^{ -5 }{ m }^{ 3 } } =2690{ kg \ m }^{ -3 }\)
Hence, the specific gravity of the sphere
\(=\frac { 2690 }{ 1000 } =2.69\)
The water in the tumbler is at a higher temperature than the bucket of normal water. But the bucket of normal water has larger internal energy than the hot water in the tumbler. This is because the internal energy is an extensive variable and it depends on the size or mass of the system.
Even though the bucket of normal water has larger internal energy than the tumbler of hot water, heat will flow from water in the tumbler to the water in the bucket. This is because heat flows from a body at higher temperature to the one at lower temperature and is independent of internal energy of the system.
Once the heat is transferred to an object it becomes internal energy of the object. The right way to say is ‘object has certain amount of internal energy’. Heat is one of the ways to increase the internal energy of a system.
The efficiency of heat engine is given by
\(\eta =1-\frac { { Q }_{ L } }{ { Q }_{ H } } \)
\(\eta =1-\frac { 300 }{ 500 } =1-\frac { 3 }{ 5 } \)
\(\eta \) = 1 – 0.6 = 0.4
The heat engine has 40% efficiency, implying that this heat engine converts only 40% of the input heat into work.
Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will decrease leading to violation of second law of thermodynamics.
Using equation
v = \(\omega \sqrt { { A }^{ 2 }-{ x }^{ 2 } } \Rightarrow { v }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }^{ 2 } \right) \)
Therefore, at position x1,
\({ v }_{ 1 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) \) .................(1)
Similarly, at position x2,
\({ v }_{ 2 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) \) ...................(2)
Subtrating (2) from (1), we get
\({ v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x_1 }^{ 2 } \right) -{ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) \)
\(= { \omega }^{ 2 }\left( { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) \)
\({ \omega }=\sqrt { \frac { { v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 } }{ { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } } } \Rightarrow T=2\pi \sqrt { \frac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 } }-{ { v }_{ 2 }^{ 2 } } } } \) ....................(3)
Dividing (1) and (2), we get
\(\frac { { v }_{ 1 }^{ 2 } }{ { v }_{ 2 }^{ 2 } } =\frac { { \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) }{ { \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) } \Rightarrow A=\sqrt { \frac { { { v }_{ 1 }^{ 2 }x }_{ 2 }^{ 2 }-{ v }_{ 2 }^{ 2 }{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 } }-{ { v }_{ 2 }^{ 2 } } } } \) .................(4)
Dividing equation (3) and equation (4), we have
\(\frac { T }{ A } =2\pi \sqrt { \frac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 }x }_{ 2 }^{ 2 }-{ { v }_{ 2 }^{ 2 }x }_{ 1 }^{ 2 } } } \)
For a fixed tension T and mass density μ, frequency is inversely proportional to the string length i.e.
\(f\propto\frac { 1 }{ l } \Rightarrow f=\frac { v }{ 2l } \Rightarrow l=\frac { v }{ 2f } \)
For the first length segment
\({ f }_{ 1 }=\frac { v }{ { 2l }_{ 1 } } \Rightarrow { l }_{ 1 }=\frac { v }{ { 2f }_{ 1 } } \)
For the second length segment
\({ f }_{ 2 }=\frac { v }{ { 2l }_{ 2 } } \Rightarrow { l }_{ 2 }=\frac { v }{ { 2f }_{ 2 } } \)
For the third length segment
\({ f }_{ 2 }=\frac { v }{ { 2l }_{ 3 } } \Rightarrow { l }_{ 3 }=\frac { v }{ { 2f }_{ 3 } } \)
Therefore, the total length
l = l1 + l2 + l3
\(\frac { v }{ 2f } =\frac { v }{ { 2f }_{ 1 } } +\frac { 2 }{ 2{ f }_{ 2 } } +\frac { v }{ 2{ f } } \Rightarrow \frac { 1 }{ f } =\frac { 1 }{ { f }_{ 1 } } +\frac { 1 }{ { f }_{ 2 } } +\frac { 1 }{ { f }_{ 3 } } \)
Physical quantities which possess dimensions and have constant values are called dimensional constants. Examples are Gravitational constant, Planck's constant etc
\({{1\ AU}\over{1\ parasec}}={{1.496\times{10}^{11}}\over{3.08\times {10}^{16}}}\)
\(={{1.496\times{10}^{11}\times{10}^{-16}}\over{3.08}}\)
\(={{1.496\times{10}^{-5}}\over{3.08}}\)
= 0.485\(\times\)10-5
= 4.85\(\times\)10-6 par sec.
4.85\(\times\)10-6 parasec present in one astronminal unit.
Technology is the application of the principles of physics for practical purposes. The application of knowledge for practical purposes in various fields to invent and produce useful products or to solve problems is known as technology. Thus, physics and technology can both together impact our society directly or indirectly. Eg: Basic laws of electricity and magnetism led to the discovery of wireless communication technology which has shrunk the world with effective communication over large distances.
\(\vec { F } =(4\vec { i } -8\vec { j } +8\vec { k } )\)
\(\left| \vec { F } \right| =\sqrt { { 4 }^{ 2 }+(-8)^{ 2 }+{ 8 }^{ 2 } } \)
\(=\sqrt { 16+64+64 } \)
\(=\sqrt { 144 } \)
=12 kg
Mass of body m= \(\frac { \left| \vec { F } \right| }{ a } =\frac { 12 }{ 2 } =6kg\)
Non-inertial frames of reference are the frames of reference in which if an object experience force, depending upon the acceleration of the frame.
\(\overrightarrow{\mathrm{F}}=8 \overline{\mathrm{i}}-10 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}\)
Mass of the body \(m=\frac{|\vec{F}|}{a}\)
\(\therefore \mathrm{m} =\frac{\sqrt{(8)^{2}+(10)^{2}+(6)^{2}}}{2}
\)
\(=\frac{\sqrt{64+100+36}}{2}
\)
\(=\frac{\sqrt{200}}{2}=\frac{\sqrt{2} \times \sqrt{100}}{2}
\)
\(=\frac{10 \sqrt{2}}{2}=5 \sqrt{2}
\)
\(=5 \times 1.414
\)
\(\text {Mass } \mathrm{m} =7.07 \mathrm{~kg}\)
\(w=\bar{F}.\bar{a}\)
\(=(-2\hat{i}+15\hat{j}+6\hat{k})-10\hat{j}\)
= 0 + 15\(\times\)10 + 0
= 150 Joule
Yes, there is ail internal energy in a body due to the thermal agitation of the particles of the body, while the vector sum of the momenta of the moving particles may be zero.
m1 = 9000 kg, u1 = 36 km/h = 10 m/s
m2 = 9000 kg, u2 = 0, v = v1= v2 =?
By conservation of momentum:
m1u1 + m2u2 = (m1 + m2)v
\(\therefore\) v = 5 m/s
Total K.E. before collision = \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }=45000J\)
Total K.E. after collision = \(\frac { 1 }{ 2 } \left( { m }_{ 1 }+{ m }_{ 2 } \right) { v }^{ 2 }=225000J\)
As total K.E. after collision < Total K.E. before the collision
\(\therefore\) Collision is inelastic.
30% of Power=\(\frac{W}{t}=\frac{mgh}{t}=\frac{V\rho gh}{t}\)
\(\frac{30}{100}\times P=\frac{V\rho gh}{t}\)
P=43.6 KW
For maximum transfer of kinetic energy during a collision, following conditions should be fulfilled.
1. The target body should be at rest.
2. The collision should be perfectly clastic.
3. The collision should be a head on collision.
4. The mass of the striking body and the target body must be exactly same.