A blue sign? that that says what is on that exit. like nearby rest area, gas station, fast food, hotel, etc
<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>
First you need to find total mechanical energy which is kinetic energy+potential energy, or (1/2mv^2+mgh). Since there is no height, we can pretty much ignore the potential energy and just calculate the kinetic.
1/2mv^2=1/2(64)(3.4)^2= 369.92J. Since it says the speed is 0 at the end of the run, we can assume that all the mechanical energy was lost. Therefore, 369.92 J of energy was lost
Answer:
7.1 m
Explanation:
Given:
Distance traveled by the student in the first attempt = 
Distance traveled by the student in the second attempt = 
So, the maximum distance that the student could travel in this attempt = 
So, the maximum distance that the student could travel in this attempt = 
Since the student first moves straight in a particular direction, rests for a while and then moves some distance in the same direction.
So, the largest distance that the student could possibly be from the starting point would be the largest distance of the final position of the student from the starting point.
And this distance is equal to the sum of the maximum distance possible in the first attempt and the second attempt of walking which is 7.1 m.
Hence, the largest distance that the student could possibly be from the starting point is 7.1 m.
