This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to the questions: Question 1.1 Concept: Newton's Second Law of Motion states that the net force acting on an object is equal to the rate of change of its momentum. Formula: F = ( p)/( t) Explanation: This formula directly defines force as the rate of change of momentum. The correct option is D. D Force Question 1.2 Concept: For an object in free fall (ignoring air friction), the acceleration due to gravity (g) is constant. Momentum (p) is the product of mass (m) and velocity (v). Step 1: Determine velocity at time t. Since the object is dropped, its initial velocity u = 0. Using the kinematic equation v = u + at, where a = g: v_t = 0 + gt = gt Step 2: Express momentum at time t. p = mv_t = mgt Step 3: Determine velocity at time 2t. v_2t = 0 + g(2t) = 2gt Step 4: Express momentum at time 2t. Let the momentum at time 2t be p'. p' = mv_2t = m(2gt) p' = 2(mgt) Step 5: Relate p' to p. Since p = mgt, we can substitute this into the expression for p': p' = 2p The correct option is C. C 2p Question 1.3 Concept: According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. If the boy exerts a force on the box, the box exerts an equal and opposite force on the boy. Explanation: The boy pushes the box with a force of 250 N. Therefore, the box exerts a force of 250 N back on the boy. The frictional force of 50 N acts on the box, opposing its motion, but it is not part of the action-reaction pair between the boy and the box. The correct option is C. C 250 N Question 1.4 Concept: Positional isomers* are compounds that have the same molecular formula and the same carbon skeleton, but differ in the position of a functional group, substituent, or multiple bond. Step 1: Determine the molecular formula and identify the carbon skeleton for each compound. I: CH C - CH_2 - CH_3 Molecular formula: C_4H_6 Name: But-1-yne (4-carbon chain, triple bond at position 1) II: CH_3 - C C - CH_2 - CH_3 Molecular formula: C_5H_8 Name: Pent-2-yne (5-carbon chain, triple bond at position 2) III: CH_3 - C C - CH_3 Molecular formula: C_4H_6 Name: But-2-yne (4-carbon chain, triple bond at position 2) IV: CH_3 - C CH Molecular formula: C_3H_4 Name: Propyne (3-carbon chain, triple bond at position 1) Step 2: Identify compounds with the same molecular formula and carbon skeleton but different positions of the multiple bond. Compounds I and III both have the molecular formula C_4H_6 and a 4-carbon chain (butyne). In compound I, the triple bond is between the first and second carbon atoms (But-1-yne). In compound III, the triple bond is between the second and third carbon atoms (But-2-yne). They differ only in the position of the triple bond. Therefore, compounds I and III are positional isomers. I and III