CBSE > Class 11 > Physics | CBSE Board | Homework Help

The ratio of output by input. You may see it at many places as in like transformers (output voltage / Input voltage), transistors, heat engines, refrigerators, and so many more.

0Thank You

ANSWER

Check the correctness of S=ut×1/2at²

Posted by Madan Mishra 2 years, 3 months ago

CBSE > Class 11 > Physics

3 answers

Anjan Karthi 2 years, 3 months ago

"S" has dimension [L]. "ut" has dimension [L/T] × [T] = [L]. "at^2" has dimension [LT^-2] × [T^2] = [L]. Thus, the equation is dimensionally correct.

1Thank You

Abhay Rana 2 years, 3 months ago

Right

0Thank You

Kaushal Ray 2 years, 3 months ago

S= ut + 1/2at²

0Thank You

ANSWER

From the top of a tower a stone is thrown with velocity 20m/s at an angle of 30degree with the horizontal the stone is strike on the ground after 4 sec find the height of the tower and horizontal range of the stone

Posted by Kartikey Gupta 2 years, 3 months ago

CBSE > Class 11 > Physics

2 answers

Anjan Karthi 2 years, 3 months ago

Wait, Range = 20 × 4 = 80m.//

0Thank You

Anjan Karthi 2 years, 3 months ago

Now, in order to reac the ground after projection upwards from a tower, we have to calculate the time for it to reach back the same height level where it was earlier (T1) and the time taken to go from that level to the ground (T2). T = T1 + T2 = 2usinπ /g + √2h/g= 2 × 20 × sin 30°/ 10 + √2h/10 = 4 sec. 2 + √2h/10 = 4. √2h/10 = 2. (2h/10 = 2^2 = 4). 2h = 40. Therefore, h = 20m.// Range R = u√2h/g = 20 × 2 = 40m.//

0Thank You

ANSWER

What is the difference between friction and frictional force?

Posted by Pooja Gupta 2 years, 3 months ago

CBSE > Class 11 > Physics

0 answers

ANSWER

A particle move with a velocity 'v' on the circumference of circle of radius 'r' after 4πr/v, it's is displacement will be ?

Posted by Rewati Raman 🤓🤓🤓 2 years, 3 months ago

CBSE > Class 11 > Physics

1 answers

Anjan Karthi 2 years, 3 months ago

s = ut + 0.5at^2 = (v × 4πr/v) + (0.5 × v^2/r × 16π^2r^2/v^2) = 4πr + 8π^2 r.//

2Thank You

ANSWER

Can you explain me integration

Posted by Shambhu Raju 2 years, 3 months ago

CBSE > Class 11 > Physics

1 answers

Saurabh Dwivedi 2 years, 3 months ago

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

Few rules to keep homework help section safe, clean and informative.

Don't post personal information, mobile numbers and other details.
Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
Be nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.
Ask specific question which are clear and concise.

Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

1Thank You

ANSWER

If p+q=r and p=q=r then angle between p and q is

Posted by Pukar Sharma 2 years, 3 months ago

CBSE > Class 11 > Physics

1 answers

Anjan Karthi 2 years, 3 months ago

R = √p^2 + q^2 + 2pq cosπ. p^2 = p^2 + p^2 + 2p^2 cosπ. -p^2 = 2p^2 cosπ. cosπ = -1/2. π = 120°.//

0Thank You

ANSWER

If the block is displaced by 20m then find work done by frictional force

Posted by Shaikh Fatema 2 years, 3 months ago

CBSE > Class 11 > Physics

0 answers

ANSWER

Important examples and exercise questions of chapter laws of motion?

Posted by Khushi Patel 2 years, 3 months ago

CBSE > Class 11 > Physics

1 answers

Ritesh Singh 2 years, 3 months ago

Newton's 2nd law of motion is the real law of motion is most important....

1Thank You

ANSWER

The sign of work done by a force on a body

Posted by Vivek Prajapat 2 years, 3 months ago

CBSE > Class 11 > Physics

1 answers

Vidhika Singhal 2 years, 3 months ago

A work is said to be done when there is displacement by the body on which the is applied

0Thank You

ANSWER

Ch 6 ncert solutions

Posted by Vivek Prajapat 2 years, 3 months ago

CBSE > Class 11 > Physics

0 answers

ANSWER

xi) The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Posted by Vibhali Mhadeshwar 2 years, 4 months ago

CBSE > Class 11 > Physics

0 answers

ANSWER

State and prove parallelogram law of vector addition

Posted by Mahak Chauhan 2 years, 4 months ago

CBSE > Class 11 > Physics

0 answers

ANSWER

Dirve a centripetal force

Posted by Rewati Raman 🤓🤓🤓 2 years, 4 months ago

CBSE > Class 11 > Physics

1 answers

Pukar Sharma 2 years, 3 months ago

Mv²/r

0Thank You

ANSWER

What is the Dimension of Energy ?

Posted by Uppu Jaideer 2 years, 4 months ago

CBSE > Class 11 > Physics

2 answers

Anjan Karthi 2 years, 4 months ago

0Thank You

Anjan Karthi 2 years, 4 months ago

Energy = [ML^2T^-2]

4Thank You

ANSWER

Establish the following vector inequalities geometrically or otherwise: b)|a+b|≥||a|-|b||

Posted by Kishore R 2 years, 4 months ago

CBSE > Class 11 > Physics

0 answers

ANSWER

Compacta module 1

Posted by Riya Mishra 2 years, 4 months ago

CBSE > Class 11 > Physics

1 answers

Shubham Yadav 2 years, 3 months ago

Don't post personal information, mobile numbers and other details.
Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
Be nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.
Ask specific question which are clear and concise.

Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

0Thank You

ANSWER

Explain potential energy

Posted by Arsh Dubey 2 years, 4 months ago

CBSE > Class 11 > Physics

4 answers

Hira Ansari 2 years, 4 months ago

Potential energy can be defined as the energy occur in the body/object due to its configuration or height. Formula for P.E.=mgh Where m=mass , g=acceleration due to gravity and h=height

0Thank You

Lakshay Tyagi 2 years, 4 months ago

Mgh

0Thank You

Anjan Karthi 2 years, 4 months ago

Hope this answers your question.

0Thank You

Anjan Karthi 2 years, 4 months ago

Potential energy is a result of the state of the body in a relative position with respect to a frame of reference (where, x = 0.) So, it is expressed generally as a scalar product of the force of interactions influencing the body and the displacement that resulted from the interactive forces, ie U = F. x. It is also associated with equilibrium, as minimum potential energy (which may be negative, meaning the forces are attractive) places body at equilibrium and higher the potential energy, less the body can be at equilibrium state.

0Thank You

ANSWER

The kepler's law period: T²=kx³, the constant k=10`³ s²m`³. Express the constant k in days and kilometre. The moon is at a distance of 3.84×10⁵km from earth obtain it's time-period of revolution in days???

Posted by Tanishq Maheshwari 2 years, 4 months ago

CBSE > Class 11 > Physics

1 answers

Anjan Karthi 2 years, 4 months ago

1 day corresponds to (24×60×60) seconds and 1 km to 1000 m. Therefore, 10^-3 s^2 / m^3 corresponds to 10^-3 × (10^3)^3 day^2 / (24 × 60 × 60)^2 km^3 = 10^6 / 7.46 × 10^9 = 10 × 10^-4 / 7.46 = 1.34 × 10^-4 day^2 / km^3 //

0Thank You

ANSWER

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg. (a) just after it is dropped from the window of a stationary train. (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h. (e) just after it is dropped from the window of a train accelerating with 1 ms, (d) lying on the floor of a train which is accelerating with 1 m s2, the stone being at rest relative to the train. Neglect air resistance throughout.

Posted by K.Kishore K.Kishore 2 years, 4 months ago

CBSE > Class 11 > Physics

1 answers

Tanishq Maheshwari 2 years, 4 months ago

(a) Given: The mass of the stone is 0.1 kg. Since the train is stationary, only the acceleration due to gravity acts in the downward direction. Newton’s second law of motion is given as, F=mg By substituting the given values in the above expression, we get F=( 0.1 kg )( 10 m/ s 2 ) =1 N Thus, a force of 1 N is acting in the downward direction. (b) Since the train is moving at constant velocity, the stone acquires only horizontal component of velocity. There is no acceleration in the horizontal direction. Only the acceleration due to gravity acts in the downward direction. Newton’s second law of motion is given as, F=mg By substituting the given values in the above expression, we get F=( 0.1 kg )( 10 m/ s 2 ) =1 N Thus, a force of 1 N is acting in the downward direction. (c) Given: The mass of the stone is 0.1 kg and the acceleration of the train is 1 m/ s 2 . As the stone is resting on the floor of the train, it experience the same acceleration as that of the train. The horizontal force acts on the stone as long as it is in the train. The horizontal force on the stone becomes zero as soon as it is dropped from the train. This is because the force at an instant doesn’t depend on the previous situations but only the situation present at that instant. Therefore, only gravitational force acts on the stone. Newton’s second law of motion is given as, F=mg By substituting the given values in the above expression, we get F=( 0.1 kg )( 10 m/ s 2 ) =1 N Thus, a force of 1 N is acting in the downward direction. (d) As the stone is at rest on the floor, it experiences the same acceleration as that of the train. The weight of the stone is balanced by the normal force applied by the floor. The force on the stone is given as, F=ma By substituting the given values in the above expression, we get F=( 0.1 kg )( 1 m/ s 2 ) =0.1 N Thus, the stone experiences force of 0.1 N in the direction of motion of the train.

0Thank You

ANSWER

A calorie is a unit of heat or energy and it equals about 4.2j where 1j=1kgm^2s^2. Suppose we emply a system of units in which the unit of mass equals akg, the unit of length equals b m, the unit of time is ys. Show that a calorie has a magnitude 4.2 a^-1 b^-2y^2 in terms of the new units.

Posted by Tanisha Kumari 2 years, 4 months ago

CBSE > Class 11 > Physics

1 answers

Tanishq Maheshwari 2 years, 4 months ago

We know that : n1u1=n2u2 n2=n1u1u2=n1[Ma1Lb1Tc1][Ma2Lb2Tc2] =n1[M1M2]a[L1L2]b[T1T2]c SI System New system n1=4.2 n2=? M1=1 kg M2=α kg L1=1 m L2=β m T1=1 s T2=γ s 1 cal=4.2 J=4.2 kg m2s−2 ∴a=1,b=2,c=−2 ∴n2=4.2[1 kgα kg]1[1 mβ m]2[1 sγ s]−2 ∴n2=4.2α−1β−2γ2 ∴1 cal=4.2α−1β−2γ2 in new system

0Thank You

ANSWER

A force F = 2i+3j+5k displaces body along 5 m along z axis. What is the work done?

Posted by Sugandha Monga 2 years, 4 months ago

CBSE > Class 11 > Physics

1 answers

Anjan Karthi 2 years, 4 months ago

W = F along z-axis (k) . Displacement along x axis = 5k.5k = 25(1) = 25J.//

0Thank You

ANSWER

Vector A= 3i-4j+5k Vector B= -2i+j+3k

Posted by Prachi Kharanshu 2 years, 4 months ago

CBSE > Class 11 > Physics

2 answers

Anjan Karthi 2 years, 4 months ago

If you are asking vector or cross product, it will be vectorically represented as (-4×3 + 5×1)i - (3×3 + -2×5)j + (3×1 + -4×-2)k = -7i + j - 5k and it's magnitude will be √(7^2 + 1^2 + 5^2) = √(49 + 1 + 25) = √75 = 5√3. //

0Thank You

Anjan Karthi 2 years, 4 months ago

If you are asking the scalar or dot product, it will be of magnitude 3i.-2i + 1j.-4j + 5k.3k = 3(-2) + 1(-4) + 5(3) = -6 - 4 + 15 = 5 units// and direction will be cos x = A.B / magnitude of A × magnitude of B = 5 / √(3^2 + 4^2 + 5^2) × √(2^2 + 1^2 + 3^2) = 5 / √50 × √14 = 5 / √700 = 5/10√7 = 1/2√7.//

0Thank You

ANSWER

A particle is thrown upwards from ground. It experiences a constant resistance forces produce a retardation of 2 ms. The ratio of time of ascent time of descent is (g=10m/s)

Posted by Harsh Gupta 2 years, 4 months ago

CBSE > Class 11 > Physics