A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. What is conserved?
(A) Momentum alone.
(B) K.E. alone.
(C) Momentum and K.E. both.
(D) P.E. alone.
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A man throws a die until he gets a number greater than `3`. The probability that he gets `5` in the last throw
Correct Answer - A `(a)` At event `A` he gets a number greater than `3` `:.P(A)=1//2` At event `B` he gets `5` in last throw `:.P(B)=1//6` `:.` Required probability is `P(E)=(1)/(6)+(1)/(2)*(1)/(6)+(1)/(2)*(1)/(2)*(1)/(6)+…..oo`...
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The rough surface of a horizontal table offers a definite maximum opposing force to initiate the motion of a block along with the table,
(A) Component of F2 adds up to weight to increase the normal reaction.
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A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. What is conserved?
Correct option is: (A) Momentum alone. When a bullet gets embbed in a solid block the whole kinetic energy gets lost . This will be perfectly inelastic c ol lision in which...
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In case of elaine collision, which s (A) Momentum and K.E. is conserved. (B) Momentum conserved and K.E. not conserved,
Correct option is: (A) Momentum and K.E. is conserved.
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How many s – block and p block elements are there in the third period of the periodic table ?
Correct option is D) 2, 6
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The coefficient of static friction between a block of mass 0.25 kg and a horizontal surface is 0.4. Find the horizontal force applied to it.
Given: μS = 0.4, m = 0.25 kg, g = 9.8 m/s2 To find: Force (FL) Formula: FL = μSN = μS(mg) Calculation: From formula, FL = 0.4 × 0.25 × 9.8 = 0.98 N The horizontal force applied...
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Calculate the force required to move a block of mass 20 kg resting on a horizontal surface, if μs = 0.3 and g = 9.8 m/s^2.
Given: m = 20 kg, μS = 0.3, g = 9.8 m/s2 To find: Force required (F) Formula: FS = μSN = μSmg Calculation: From formula, FS = 3 × 20 × 9.8 = 58.8N The force required to move...
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A block of mass 40 kg resting on a horizontal surface just begins to slide when a horizontal force of 120 N is applied to it.
Given: FL = 120N, Fk = 80N, m = 40 kg, g = 9.8 m/s2 To find: i. Coefficient of static friction (μS) ii. Coefficient of kinetic friction (μk) Formulae: i. μs = \(\frac{F_L}{N}\) ii. μk = \(\frac{F_K}{N}\) Calculation: From formula (i) we get. N =...
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A 20 kg metal block is placed on a horizontal surface. The block just begins to slide when horizontal force of 100 N is applied to it.
Given: m = 20 kg, FL = 100 N, μk = 0.4 To find: i. Coefficient of static friction (μS) ii. Minimum force required (Fk) Formulae: i. μs = \(\frac{F_L}{N}=\frac{F_L}{mg}\) ii. μS = \(\frac{F_K}{N}\) Calculation: From formula (i), μs = \(\frac{100}{20\times9.8}\) = 0.5102 From formula (ii), Fk = μkN...
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A wooden block of 50 kg is at rest on the table. A force of 70 N is required to just slide the block. The coefficient of static friction is
Correct option is: (D) \(\frac{1}{7}\)
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