The solution to the problem is as follows:
<span>First, I'd convert 188 mi/hr to ft/s. You should end up with about ~275.7 ft/s.
So now write down all the values you know:
Vfinal = 275.7 ft/s
Vinitial = 0 ft/s
distance = 299ft
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<span>Now just plug in Vf, Vi and d to solve
</span>
<span>Vf^2 = Vi^2 + 2 a d
</span><span>BTW: That will give you the acceleration in ft/s^2. You can convert that to "g"s by dividing it by 32 since 1 g is 32 ft/s^2.</span>
Answer: a) Mr = 2.4×10^-4kg/s
V = 34.42m/a
b) E = 173J
Ø = 2693.1J
c) Er = 0.64J/s
Explanation: Please find the attached file for the solution
Answer:
The difference between the velocity graph made walking at a steady rate means that its the same value in time, that means there's no slope on the graph, so its acceleration is 0
On the other hand, if the velocity is increasing with time, the slope of the graph becomes positive, which means that the acceleration of the particle is positive.
Answer:
The magnitude of the change in momentum of the stone is 5.51kg*m/s.
Explanation:
the final kinetic energy = 1/2(0.15)v^2
1/2(0.15)v^2 = 70%*1/2(0.15)(20)^2
v^2 = 21/0.075
v^2 = 280
v = 16.73 m.s
if u is the initial speed and v is the final speed, then:
u = 20 m/s and v = - 16.73m/s
change in momentum = m(v-u)
= 0.15(- 16.73-20)
= -5.51 kg*m.s
Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.
