Answer
given,
x = (3.9 cm)sin[(9.3 rad/s)πt]
general equation of displacement
x = A sin ω t
A is amplitude
now on comparing
c) Amplitude =3.9 cm
a) frequency =
f = 4.65 Hz
b) period of motion
T = 0.215 s
d) time when displacement is equal to x= 2.6 cm
x = (3.9 cm)sin[(9.3 rad/s)πt]
2.6 = (3.9 cm)sin[(9.3 rad/s)πt]
sin[(9.3 rad/s)πt] = 0.667
9.3 π t = 0.73
t = 0.025 s
Refer to the diagram shown below.
The force, F, is applied at 5 cm from the elbow.
For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N
Answer: 1500 N
Answer:
w = 25.05 rad / s
, α = 0.7807 rad / s²
, θ = 1972.75
Explanation:
This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off
θ = w₀ t - ½ α t²
α = (w₀t - θ) 2/t²
let's reduce the magnitudes to the SI system
w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s
θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad
let's calculate the angular acceleration
α = (55.5 39 - 1570.8) 2/39²
α = 0.7807 rad / s²
having the acceleration we can calculate the final speed
w = w₀ - ∝ t
w = 55.5 - 0.7807 39
w = 25.05 rad / s
the time to stop w = 0
0 = wo - alpha t
t = wo / alpha
t = 55.5 / 0.7807
t = 71.09 s
the angle traveled
w² = w₀⁹ - 2 α θ
w = 0
θ = w₀² / 2α
let's calculate
θ = 55.5 2 / (2 0.7807)
θ = 1972.75
D
Using the kinetic energy 1/2mv^2 formula
5*10^5 is the answer
Gravity has an effect on weight, but not on mass. The mass of the object will be the same on both planets. (choice-C)
