Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
Answer:
V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2
V = 73.8 m/s
tan theta = Vy / Vx = 62/40 = 1.55
theta = 57.2 deg
Answer:
the mean
Explanation:
The mean (informally, the “average“) is found by adding all of the numbers together and dividing by the number of items in the set: 10 + 10 + 20 + 40 + 70 / 5 = 30. The median is found by ordering the set from lowest to highest and finding the exact middle. The median is just the middle number: 20.
Answer:
Vprom = 0.00347[km/min]
Explanation:
We can calculate each of the average speeds and then perform the overall average between the two speeds.
V1 = 6/54
V1 = 0.111[km/min]
V2 = 1/16
V2 = 0.0625[km/min]
![V_{prom} = \frac{V_{1} + V_{2}}{2} \\V_{prom} = \frac{0.1111 + 0.0625}{2}\\V_{prom} = 0.00347 [km/min]](https://tex.z-dn.net/?f=V_%7Bprom%7D%20%3D%20%5Cfrac%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%7B2%7D%20%20%5C%5CV_%7Bprom%7D%20%3D%20%5Cfrac%7B0.1111%20%2B%200.0625%7D%7B2%7D%5C%5CV_%7Bprom%7D%20%3D%200.00347%20%5Bkm%2Fmin%5D)
