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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.

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Check the correctness of S=ut×1/2at²

Posted by Madan Mishra 1 year, 11 months ago

CBSE > Class 11 > Physics

3 answers

Anjan Karthi 1 year, 11 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.

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Abhay Rana 1 year, 11 months ago

Right

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Kaushal Ray 1 year, 11 months ago

S= ut + 1/2at²

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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 1 year, 11 months ago

CBSE > Class 11 > Physics

2 answers

Anjan Karthi 1 year, 11 months ago

Wait, Range = 20 × 4 = 80m.//

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Anjan Karthi 1 year, 11 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.//

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What is the difference between friction and frictional force?

Posted by Pooja Gupta 1 year, 11 months ago

CBSE > Class 11 > Physics

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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 ago

CBSE > Class 11 > Physics

1 answers

Anjan Karthi 2 years ago

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

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Can you explain me integration

Posted by Shambhu Raju 2 years ago

CBSE > Class 11 > Physics

1 answers

Saurabh Dwivedi 1 year, 11 months ago

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

Posted by Pukar Sharma 2 years ago

CBSE > Class 11 > Physics

1 answers

Anjan Karthi 2 years 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°.//

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

Posted by Shaikh Fatema 2 years ago

CBSE > Class 11 > Physics

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Important examples and exercise questions of chapter laws of motion?

Posted by Khushi Patel 2 years ago

CBSE > Class 11 > Physics

1 answers

Ritesh Singh 1 year, 11 months ago

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

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The sign of work done by a force on a body

Posted by Vivek Prajapat 2 years ago

CBSE > Class 11 > Physics

1 answers

Vidhika Singhal 2 years ago

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

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Ch 6 ncert solutions

Posted by Vivek Prajapat 2 years ago

CBSE > Class 11 > Physics

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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 ago

CBSE > Class 11 > Physics

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State and prove parallelogram law of vector addition

Posted by Aaditya Chauhan 2 years ago

CBSE > Class 11 > Physics

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Dirve a centripetal force

Posted by Rewati Raman 🤓🤓🤓 2 years ago

CBSE > Class 11 > Physics

1 answers

Pukar Sharma 2 years ago

Mv²/r

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What is the Dimension of Energy ?

Posted by Uppu Jaideer 2 years ago

CBSE > Class 11 > Physics

2 answers

Anjan Karthi 2 years ago

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Anjan Karthi 2 years ago

Energy = [ML^2T^-2]

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

Posted by Kishore R 2 years ago

CBSE > Class 11 > Physics

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Compacta module 1

Posted by Riya Mishra 2 years ago

CBSE > Class 11 > Physics

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Shubham Yadav 1 year, 11 months ago

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Explain potential energy

Posted by Arsh Dubey 2 years ago

CBSE > Class 11 > Physics

4 answers

Hira Ansari 2 years 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

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Lakshay Tyagi 2 years ago

Mgh

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Anjan Karthi 2 years ago

Hope this answers your question.

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Anjan Karthi 2 years 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.

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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 ago

CBSE > Class 11 > Physics

1 answers

Anjan Karthi 2 years 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 //

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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 ago

CBSE > Class 11 > Physics

1 answers

Tanishq Maheshwari 2 years 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.

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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 ago

CBSE > Class 11 > Physics

1 answers

Tanishq Maheshwari 2 years 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

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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 ago

CBSE > Class 11 > Physics

1 answers

Anjan Karthi 2 years ago

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

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

Posted by Prachi Kharanshu 2 years ago

CBSE > Class 11 > Physics

2 answers

Anjan Karthi 2 years 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. //

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Anjan Karthi 2 years 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.//

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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 ago

CBSE > Class 11 > Physics