Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.
From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.
Hence P(t)=a*t^2/2+10t
Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T
Hence T=225/30=7.5
It took 7.5 seconds
Answer:
H = 109.2 TJ.
Explanation:
Please see the attachment below.
Answer:
26.5 m
Explanation:
= initial position of the object = 75.2 m
= final position of the object
= displacement of the object = - 48.7
Displacement of the object is given as the difference of final and initial position of the object
Inserting the values
- 48.7 = x - 75.2
x = 26.5 m
B) 14.0 N
The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is:
E = 0.5 M V^2
where
E = Energy
M = Mass
V = velocity
Substituting the known values, let's calculate the before and after energy.
Before:
E = 0.5 M V^2
E = 0.5 13.5kg (2.25 m/s)^2
E = 6.75 kg 5.0625 m^2/s^2
E = 34.17188 kg*m^2/s^2 = 34.17188 joules
After:
E = 0.5 M V^2
E = 0.5 13.5kg (1.2 m/s)^2
E = 6.75 kg 1.44 m^2/s^2
E = 9.72 kg*m^2/s^2 = 9.72 Joules
So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N
Rounding to 1 decimal place gives 14.0 N which matches option "B".
