In a laboratory model of cars skidding to a stop, data are measured for four trials using two blocks. The blocks have identical masses but different coefficients of kinetic friction with a table: µ k = 0.2 and 0.8. Each block is launched with speed v i = 1 m/s and slides across the level table as the block comes to rest. This process represents the first two trials. For the next two trials, the procedure is repeated but the blocks are launched with speed v i = 2 m/s. Rank the four trials (a) through (d) according to the stopping distance from largest to smallest. If the stopping distance is the same in two cases, give them equal rank. (a) v i = 1 m/s, = 0.2 (b) v i = 1 m/s, µ k = 0.8 (c) v i = 2 m/s, = 0.2 (d) v i =2 m/s, µ k = 0.8
In a laboratory model of cars skidding to a stop, data are measured for four trials using two blocks. The blocks have identical masses but different coefficients of kinetic friction with a table: µ k = 0.2 and 0.8. Each block is launched with speed v i = 1 m/s and slides across the level table as the block comes to rest. This process represents the first two trials. For the next two trials, the procedure is repeated but the blocks are launched with speed v i = 2 m/s. Rank the four trials (a) through (d) according to the stopping distance from largest to smallest. If the stopping distance is the same in two cases, give them equal rank. (a) v i = 1 m/s, = 0.2 (b) v i = 1 m/s, µ k = 0.8 (c) v i = 2 m/s, = 0.2 (d) v i =2 m/s, µ k = 0.8
Solution Summary: The author explains how the rank of four trials based on the stopping distance is c>a=d>b.
In a laboratory model of cars skidding to a stop, data are measured for four trials using two blocks. The blocks have identical masses but different coefficients of kinetic friction with a table: µk = 0.2 and 0.8. Each block is launched with speed vi = 1 m/s and slides across the level table as the block comes to rest. This process represents the first two trials. For the next two trials, the procedure is repeated but the blocks are launched with speed vi = 2 m/s. Rank the four trials (a) through (d) according to the stopping distance from largest to smallest. If the stopping distance is the same in two cases, give them equal rank. (a) vi = 1 m/s, = 0.2 (b) vi = 1 m/s, µk = 0.8 (c) vi = 2 m/s, = 0.2 (d) vi =2 m/s, µk = 0.8
the cable may break and cause severe injury.
cable is more likely to break as compared to the
[1]
ds, inclined at angles of 30° and 50° to the vertical
rings by way of a scaled diagram. [4]
I
30°
T₁
3cm
3.8T2
cm
200 N
50°
at it is headed due North and its airspeed indicat
240 km/h. If there is a wind of 100 km/h from We
e relative to the Earth? [3]
Can you explain this using nodal analysis
With the nodes I have present
And then show me how many KCL equations I need to write, I’m thinking 2 since we have 2 dependent sources
Chapter 8 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
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