Answer:
29.4m/s
Explanation:
Given parameters:
Time = 3s
Unknown:
Average velocity = ?
Solution:
To solve this problem, we use the expression below:
v = u + gt
v is the average velocity
u is the initial velocity = 0m/s
g is the acceleration due to gravity = 9.8m/s²
t is the time
So;
v = 0 + (9.8 x 3) = 29.4m/s
The model bridge captures all the structural attributes of the real bridge, at a reduced scale.
Part a.
Note that volume is proportional to the cube of length. Therefore the actual bridge will have 100^3 = 10^6 times the mass of the model bridge.
Because the model bridge weighs 50 N, the real bridge weighs
(50 N)*10^6 = 50 MN.
Part b.
The model bridge matches the structural characteristics of the actual bridge.
Therefore the real bridge will not sag either.
Answer:
Imp = 25 [kg*m/s]
v₂= 20 [m/s]
Explanation:
In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.
1)
where:
m₁ = mass of the object = 5 [kg]
v₁ = initial velocity = 0 (initially at rest)
F = force = 5 [N]
t = time = 5 [s]
v₂ = velocity after the momentum [m/s]
![(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]](https://tex.z-dn.net/?f=%285%2A0%29%20%2B%285%2A5%29%20%3D%20%28m_%7B1%7D%2Av_%7B2%7D%29%20%3D%20Imp%5C%5CImp%20%3D%2025%20%5Bkg%2Am%2Fs%5D)
2)
![(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]](https://tex.z-dn.net/?f=%28m_%7B1%7D%2Av_%7B1%7D%29%2B%28F%2At%29%3D%28m_%7B1%7D%2Av_%7B2%7D%29%5C%5C%280.075%2A0%29%2B%2830%2A0.05%29%3D%280.075%2Av_%7B2%7D%29%5C%5Cv_%7B2%7D%3D20%20%5Bm%2Fs%5D)
Answer:
speed = distance/time
Explanation:
speed = 150/30
speed =5m/s
you were running fast .....5m/s is a good speed
The correct answer would be left
