Answer:
Torque = –207.4 Nm
Explanation:
Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)
Torque = I × α
α = angular acceleration
I = moment of inertia
I = MR² for a circular hoop
Torque = 3.2×5.4×(– 12)
Torque = –207.4 Nm
Answer:
a) T = 1,467 s
, b) A = 0.495 m
, c) v = 4.97 10⁻² m / s
Explanation:
The simple harmonic movement is described by the expression
x = A cos (wt + Ф)
Where the angular velocity is
w = √ k / m
a) Ask the period
Angular velocity, frequency and period are related
w = 2π f = 2π / T
T = 2π / w
T = 2pi √ m / k
T = 2π √ (1.2 / 22)
T = 1,467 s
f = 1 / T
f = 0.68 Hz
b) ask the amplitude
The mechanical energy of a harmonic oscillator
E = ½ k A²
A = √2 E / k
A = √ (2 2.7 / 22)
A = 0.495 m
c) the mass changes to 8.0 kg
As released from rest Ф = 0, the equation remains
x = A cos wt
w = √ (22/8)
w = 1,658
x = 3.0 cos (1,658 t)
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum when without wt = ±1
v = Aw
v = 0.03 1,658
v = 4.97 10⁻² m / s
Answer:
In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers
Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
Answer: 50 m/s
Explanation: speed v = 2· pi·n·r = 2· 3.14· 2 s^-1· 4 m
