Answer:
a) She get detention for being late again
Explanation:
First, we need to identify how much time does she take on each hallway.
With the distance and the speed, we can calculate the time as:
t = distance/speed
So, for each hallway, we get:
First hallway:
distance = 35 m
speed = 3.5 m/s
time = 35/3.5 = 10s
Second Hallway
distance = 48 m
speed = 1.2 m/s
time = 40s
Third Hallway
distance = 60 m
speed = 5 m/s
time = 60/5 = 12 s
Therefore, the total time that she takes was
10s + 40s + 12s = 62s
Since she takes more than 60 seconds, she will be late again.
Finally, we know that she takes 10s to run a distance of 35m, then another 40s to run a distance of 48 m, and another 12s to run a distance of 60 m. Therefore, the distance vs. time graph for this situation is
Explanation:
We use Hooke's Law, which states that,
F=kx
F- is the force in newtons
k- is the spring constant
x- is the extension of the spring in meters
Solving for k we get k=F/x
k= 10N/0.50m
=100 N/m
Construction, like building a home/building, digging, like in a mine, and opening a soda can, where the part to open is a lever.
Answer:
The arrow will leave the bow with a velocity of 10 m/s.
Explanation:
Hi there!
The potential energy stored in the bow can be calculated using the following equation:
U = 1/2 · k · d²
Where
U = elastic potential energy.
k = spring constant.
d = stretched distance of the bow
Then:
U = 1/2 · 112 N/m · (0.29 m)²
U = 4.7 J
When the bow is released, the potential energy is transformed into kinetic energy. Then, the kinetic energy of the arrow when it leaves the bow will be:
KE = 1/2 · m · v² = 4.7J
Where:
KE = kinetic energy.
m = mass of the arrow.
v = velocity of the arrow:
Then:
4.7 J = 1/2 ·0.094 kg · v²
2 · 4.7 J / 0.094 kg = v²
9.4 kg · m²/s² / 0.094 kg = v²
v = 10 m/s
The arrow will leave the bow with a velocity of 10 m/s.