Answer:
Final velocity at the top of the ramp is 6.58m/s
Explanation
Check the attachment
Answer: 30.34m/s
Explanation:
The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg
Sum of forces in the x direction
mv²/r = N sin 28 + μN cos 28
mv²/r = N(sin 28 + μcos 28)
Thus,
mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]
v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]
v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]
v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]
v²/36 = 9.8 (1.2376/0.4744)
v²/36 = 9.8 * 2.6088
v²/36 = 25.57
v² = 920.52
v = 30.34m/s
Explanation:
Since its accelerating, the velocity vs time graph is linear
For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula
A= (vf - vi)/t
a= vf-0/t
1.25=vf / 7
1.25*7=vf
8.75 = vf
Now for displacement plug all the values in
X = 1/2(vf-vi)/t formula
The displacement (x) is 30.625 m
For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above
The answer is t = 5.02
Im pretty sure all the answers are correct
Answer:
28.81 m
Explanation:
Ff = -123
m * a = -123
(29.8+10.3) * a = -123
a = -123/40.1 = -3.07
We know,
v^2 = u^2 + 2as
0^2 = 13.3^2 + 2*(-3.07)*s
s = 176.89/6.14 = 28.81
[ If there's a problem with the solution, pleaase let me know ]
