Part 1 Select and circle the correct answer. Selecting more than 1 answers or no
ID: 1785006 • Letter: P
Question
Part 1 Select and circle the correct answer. Selecting more than 1 answers or not selecting an answer will earn zero credits. 1. Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Just as they reach the lake below A. the speed of Alice is larger than that of Tom. B. the splashdown speed of Alice is larger than that of Tom. C. they will both have the same speed. D. the speed of Tom will always be 9.8 m/s larger than that of Alice. E. the speed of Alice will always be 25 m/s larger than that of Tom. What is the maximum distance we can shoot a dart, from ground level, provided our toy dart gun gives a maximum initial velocity of 2.78 m/s and air resistance is negligible? 2. A. 0.789 m B. 1.58 nm C. 1.39 m D. 0.394 m An object moves in a circle of radius R at constant speed with a period T. If you want to change only the period in order to cut the object's acceleration in half, the new period should be 3. C. Th/2 D. 7/2 E. 4T The figure shows a graph of the acceleration of a 125-g object as a function of the net force acting on it. What is the acceleration at points A and B? 4. a) 16 m/s2 and 4.0 m/s b) 6.3 x102 m/s2 and 0.25 m/s c) 20 m/s2 and 0.5 m/s d) Not enough information to calculate the accelerations 1.0 2.0Explanation / Answer
1) Option A is correct .This is because both Tom and Alice will have equal vertical veolcity component but Alice will have horizontal component of velocity as well. This is why ,speed(magnitude of velocity) of Alice will be higher than that of Tom
2)For maximumm horizontal distance, the angle at which the dart should be shot from ground should be 45o
Therefore, the maximum horizontal distance travelled = 2.782sin90 / 9.8=0.789 m/s
3)Let the initial constant speed be v
Radius=R
Time period = T=2*pi*R/v
Acceleration =a= v2/R=(2*pi*R/T)2/R = 4*pi2R/T2
For acceleration to become half of initial acceleration, i.e., a/2, the new time period should be T*sqrt(2) . Therefore,correct option is D
4)Force at A=2 N , mass=0.125kg
acceleration at A = 2/0.125=16 m/s2
Force at B=0.5 N
acceleration at B = 0.5/0.125=4 m/s2
Therefore, correct option is (a)