Answer:
38 m/s
43 m/s
Explanation:
x = 18t + 5.0t²
The instantaneous velocity is the first derivative:
v = 18 + 10.t
At t = 2.0:
v = 18 + 10.(2.0)
v = 38 m/s
The average velocity is the change in position over change in time.
v = Δx / Δt
v = [ (18t₂ + 5.0t₂²) − (18t₁ + 5.0t₁²) ] / (t₂ − t₁)
Between t = 2.0 and t = 3.0:
v = [ (18(3.0) + 5.0(3.0)²) − (18(2.0) + 5.0(2.0)²) ] / (3.0 − 2.0)
v = [ (54 + 45) − (36 + 20.) ] / 1.0
v = 99 − 56
v = 43 m/s
Vo = 18 m/s
angle 35 degrees
1) Components of the initial velocity
Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s
2) Equations of postion:
x = Vox*t
y = Voy*t - gt^2 / 2
3) Calculations
A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s
x = 14.74 * t
t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m
t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m
t = 1.5s => x = 22.11 m
t = 2s => x = 29.48 m
B)
y = Voy*t - gt^2 / 2
Voy = 10.32 m/s
g = 10 m/s (approximation)
y = 10.32*t - 5t^2
t = 0.5 s=> y = 3.91m
t = 1 s => y = 5.32m
t = 1.5 s => y = 4.23m
t = 2 s => y = 0.64 m
Answer: This number has three significant figures.
Explanation: Trailing zeros, which are zeros at the end of a number, are significant only if the number has a decimal point. Thus, in 1,500 m, the two trailing zeros are not significant because the number is written without a decimal point.
Kinetic friction for the truck is given by:
F = μ*N = μ*m*g
μ coefficient of friction
m mass of the truck
g gravitational constant = 9.81
Answer:
0.5
Explanation:
From the question given above, the following data were obtained:
Force (F) = 50 N
Weight (W) = 100 N
Coefficient of static friction (μ) =.?
The coefficients of static friction (μ) is simply defined as the ratio of force (F) to the normal reaction (N). Mathematically, it can be expressed as:
μ = F / N
With the above formula, we can obtain the coefficient of static friction as follow:
Force (F) = 50 N
Weight (W) = 100 N
Coefficient of static friction (μ) =.?
μ = F / W
μ = 50 / 100
μ = 0.5
Thus, coefficient of static friction is 0.5
