A particle starts with an initial velocity of 200 cm/s and moves with a uniform retardation of 10 cm/s². If it describes 1500 cm in time t, what is/are the possible value(s) off? (a) 10 sec only (c) 30 sec only (b) 10 sec and 30 sec (d) 5 sec and 10 sec
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- Each measurement we take has a level of uncertainty in it. The smaller the uncertainty, the larger the precision our measurement has. Let's revisit how one might go about estimating the uncertainty in a measurement. If we assume our instrument is properly calibrated and we are not introducing systematic error through improper techniques, then if we take enough measurements, the average of these measurements will be around the true value. The upper and lower bounds of these measurements would then give us one method for determining our measuring device's uncertainty. (This of course is only true for a large set of measurements, but the approximation is good enough for now to get us started). For example, say we perform last week's experiment of pulling a block at a constant velocity with a force sensor across another rough surface. We take the following five measurements: F, = {2.51 N, 2.53 N, 2.50 N, 2.54 N, 2.49 N} The average of this dataset is F, = 2.51 N Now we have the choice of…A rectangle has the dimensions of 3.1 mx 2.40 m when viewed by someone at rest with respect to it. When you move past the rectangle along one of its sides, the rectangle looks like a square. Then you move at the same speed along the adjacent side of the rectangle. What dimensions do you observe (a) for the longer side and (b) for the shorter side? (a) Number i (b) Number Units UnitsBob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, is a baseball, a stopwatch, and a friend on the beach below with a long measuring tape. Bob is a pitcher and he knows that the fastest he can throw the ball is about vo = 34.5 m/s. Bob starts the stopwatch as he throws the ball (with no way to measure the ball's initial trajectory), and watches carefully. The ball rises and then falls, and after t₁ = 0.510 s the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball landed x = 128 m from the base of the cliff. How high up is Bob, if the ball started exactly 2 m above the edge of the cliff? 2 m -- X Bob's position= x10 TOOLS m
- Should be solved using classical mechanics( Kinematics) Many physicists rely on data from particle accelerators for their research. Consider an alpha particle (the nucleus of a helium atom) traveling inside a straight tube 2.0 m long that forms part of a particle accelerator. The alpha particle enters the tube moving at a velocity 9.5 × 105 m/s and emerges from the other end after 8.0 × 10−7 seconds. (a) If the particle’s acceleration is constant, what is its velocity when it leaves the tube? (b) If instead the particle’s acceleration increases linearly with time as a(t) = kt, where k = 1 × 1019 m/s3, what is its velocity when it leaves the tube? (c) Plot the velocity versus time graph for the alpha particle in parts (a) and (b), for the duration of its travel in the tube.In the answer, it gives a formula f-ff=m(0) and afterwards simplies into 120 - ff = 0 and further into ff=120N. My question is how do you not have to divide by 0 to finish the equation if that makes sense. To simply further why are we able to ignore the m(0) portion?I1:00:00 PM Problem 3: Suppose the speed of light were only 3000.0 m's. A jet fighter moving toward a target on the ground at 810 m/s shoots bullets, each having a muzzle velocity of 1050 m's. Randomized Variables v-810 m/s v2 1050 m's ed What is the velocity of the bullets relative to the target in km's? Grade Sommary Deductions 0% Potential 100%
- When measuring with a muline, it was measured as n=10 when the speed was 1.1 m/s, and as n=32 when the speed was 1.5 m/s. What is the velocity formula?Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2axv2=v02+2ax. The problem requires solving for xx, and the known values for the system are a=2.55meter/second2a=2.55meter/second2, v0=21.8meter/secondv0=21.8meter/second, and v=0meter/secondv=0meter/second. Perform the next step in the analysis.G docs.google.com & An object moves on the negative x axis a distance of (5m) according to the formula (x= -7t +2t^2), the time it * :takes is equal to the formula (X= -7t + 2t²) 2s 4s 2.5s O 3.5s