Answer:
abiotic
Explanation:
i think but dont take my word for it
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
The momentum of the three objects are as follow :
11 kg-m/s, -65 kg-m/s and -100 kg-m/s
Before collision, the momentum of the system is :
After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.
It would mean that, after collision, momentum of the system is equal to the initial momentum.
Hence, final momentum = -154 kg-m/s.
Average velocity is a vector unit (i.e. includes magnitude <em>and </em>direction) calculated by working out distance ÷ time:
80 metres ÷ 20 seconds = 4 metres/seconds (m/s)
Therefore, your final answer is C. 4 m/s south.
