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  • 2 answers

Anisha Patel 2 years, 1 month ago

9.8

Anjan Karthi 2 years, 1 month ago

G = 6.673 × 10^-11 Nm2/kg2 = 6.673 × 10^-8 dyne cm2/g2
  • 1 answers

Anjan Karthi 2 years ago

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.
  • 3 answers

Anjan Karthi 2 years, 1 month 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.

Abhay Rana 2 years, 1 month ago

Right

Kaushal Ray 2 years, 1 month ago

S= ut + 1/2at²
  • 2 answers

Anjan Karthi 2 years, 1 month ago

Wait, Range = 20 × 4 = 80m.//

Anjan Karthi 2 years, 1 month 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.//
  • 1 answers

Anjan Karthi 2 years, 2 months ago

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

Saurabh Dwivedi 2 years, 1 month ago

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  • 1 answers

Anjan Karthi 2 years, 2 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°.//
  • 1 answers

Ritesh Singh 2 years, 1 month ago

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

Vidhika Singhal 2 years, 2 months ago

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

Pukar Sharma 2 years, 2 months ago

Mv²/r
  • 2 answers

Anjan Karthi 2 years, 2 months ago

D

Anjan Karthi 2 years, 2 months ago

Energy = [ML^2T^-2]
  • 1 answers

Shubham Yadav 2 years, 1 month 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

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  • 4 answers

Hira Ansari 2 years, 2 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

Lakshay Tyagi 2 years, 2 months ago

Mgh

Anjan Karthi 2 years, 2 months ago

Hope this answers your question.

Anjan Karthi 2 years, 2 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.
  • 1 answers

Anjan Karthi 2 years, 2 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 //
  • 1 answers

Tanishq Maheshwari 2 years, 2 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.
  • 1 answers

Tanishq Maheshwari 2 years, 2 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
  • 1 answers

Anjan Karthi 2 years, 2 months ago

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

Anjan Karthi 2 years, 2 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. //

Anjan Karthi 2 years, 2 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.//
  • 1 answers

Anjan Karthi 2 years, 2 months ago

Time of ascent = √2h/(g-a). Time of descent = √2h/(g+a). Ratio = g+a/g-a = 12/8 = 3:2//
  • 1 answers

Sazia Hashmi 2 years, 3 months ago

The rate of change of angular momentum is equal to applied torque
  • 2 answers

Shaikh Fatema 2 years, 2 months ago

You can get it on google easily Type or search Physics part one notes by any classes like Unacademy Or vedantu

Bobby Arya 2 years, 2 months ago

Ok
  • 5 answers

Hira Ansari 2 years, 2 months ago

Ampere Denoted by A.

Zeeshan Ansari 2 years, 2 months ago

Watt

Sazia Hashmi 2 years, 3 months ago

Ampere

Bhavya Lenka 2 years, 3 months ago

Ampere (A)

Kartik Joshi 2 years, 3 months ago

Coulumb

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