The distance covered is 115 m
Explanation:
The motion of Ileana is a uniformly accelerated motion (constant acceleration), therefore we can use the following suvat equation:

where
s is the distance covered
u is the initiaal velocity
v is the final velocity
t is the time elapsed
In this problem, we have:
u = 4.20 m/s
v = 5.00 m/s
t = 25.0 s
Therefore, we can re-arrange the equation to find the distance covered:

Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
#LearnwithBrainly
Let the angle be Θ (theta)
Let the mass of the crate be m.
a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.
Normal force (N) = mg CosΘ
μ (coefficient of static friction) = 0.29
Static friction = μN = μmg CosΘ
Now, along the ramp, the equation of net force will be:
mg SinΘ - μmg CosΘ = 0
mg SinΘ = μmg CosΘ
tan Θ = μ
tan Θ = 0.29
Θ = 16.17°
b) Let the acceleration be a.
Coefficient of kinetic friction = μ = 0.26
Now, the equation of net force will be:
mg sinΘ - μ mg CosΘ = ma
a = g SinΘ - μg CosΘ
Plugging the values
a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96
a = 2.7244 - 2.44608
a = 0.278 m/s^2
Hence, the acceleration is 0.278 m/s^2
Hi there!

We know that:

U = Potential Energy (J)
K = Kinetic Energy (J)
E = Total Energy (J)
At 10m, the total amount of energy is equivalent to:
U + K = 50 + 50 = 100 J
To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:
100J = mgh
We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².
50J = 10(10)m
m = 1/2 kg
Now, solve for height given that E = 100 J:
100J = 1/2(10)h
100J = 5h
<u>h = 20 meters</u>
Answer:
a) 141.6m
b) 8.4m/s
Explanation:
a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

hence, the total distance is 141.6m
b) the mean velocity of the total trajectory is given by:

hence, the mean velocity is 8.4 m/s