<span>Step 1 -- determine the acceleration of the 200-g block after bullet hits it
a = (coeff of friction) * g
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
a = 0.400*9.8
a = 3.92 m/sec^2
Step 2 -- determine the speed of the block after the bullet hits it
Vf^2 - Vb^2 = 2(a)(s)
where
Vf = final velocity = 0 (since it will stop)
Vb = velocity of block after bullet hits it
a = -3.92 m/sec^2
s = stopping distance = 8 m (given)
Substituting values,
0 - Vb^2 = 2(-3.92)(8)
Vb^2 = 62.72
Vb = 7.92 m/sec.
M1V1 + M2V2 = (M1 + M2)Vb
where
M1 = mass of the bullet = 10 g (given) = 0.010 kg.
V1 = velocity of bullet before impact
M2 = mass of block = 200 g (given) = 0.2 kg.
V2 = initial velocity of block = 0
Vb = 7.92 m/sec
Substituting values,
0.010(V1) + 0.2(0) = (0.010 + 0.2)(7.92)
Solving for V1,
V1 = 166.32 m/sec.
Therefore the answer is (B) 166 m/s!</span>
Answer:
average velocity = displacement/time
A. increases with security and accuracy of stored and dispense medications
A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?
<span>Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. </span>
<span>At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. </span>
<span>So total time is 8.5 + 25.5 or 34 sec. </span>
<span>Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.</span>
Answer:
your questions are not clear
