this question is unclear. please specify.
Answer:
a = -2.82 m/s²
v = 18.4 m/s
Explanation:
Initial velocity, u = 26.0 m/s
Final velocity, v = 0 m/s
Distance travelled, s = 120m
Using
v² = u² + 2as
0 = 26² + 2(120)a
a= -2.82 m/s²
Velocity of the car when it was 60.0m past the point where the brakes were applied:
u = 26.0 m/s
a = -2.82 m/s²
s = 60m
Using
v² = u² + 2as
v² = 26² + 2(-2.82)(60)
v² = 337.6
v = 18.4 m/s
Answer:
Option D - 0.2 s
Explanation:
We are given;
Initial velocity; u = 7 m/s
Height of table; h = 1.8m
Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.
Thus;
h = ut + ½gt²
Plugging in the relevant values, we have;
1.8 = 7t + ½(9.8)t²
4.9t² + 7t - 1.8 = 0
Using quadratic formula to find the roots of the equation gives us;
t = -1.65 or 0.22
We can't have negative t value, thus we will pick the positive one.
So, t = 0.22 s
This is approximately 0.2 s
Answer:
.team 1 a = 20.8 m/s² team 2 a = 19.2 m/s²
Explanation:
Let's use Newton's second law to calculate the acceleration of the two groups
F = ma
a = F / M
Group 1
a = 9 1354/9 65
a = 20.8 m / s²
Group 2
a = 9 1364/9 71
a = 19.2 m / s²
If the two groups pull against each other, group 1 should win, by creating a greater acceleration.
Δa = 20.8 -19.2
Δa = 1.6 m / s²
Energy conservation
mgh = 1/2mv^2
1.5*9.8*3 = 1/2*1.5*v^2
44.1*2 = 1.5*v^2
88.2/1.5 = v^2
58.8 = v^2
v = 7.668 m/s
