Answer:
I believe it is 1/6 M
Explanation:
sorry if i am not right but i think i am
Answer:
1/Re=1/R1 +1/R2= 2x1/R1=2/R1=2/R2=2/50=1/25
1/Re=1/25, so the answer is Re= 25
the answer is A. 25.0
Answer:
Explanation:
Let Torque due to friction be
F
Net torque
= 46 - F
Angular impulse = change in angular momentum
=( 46 - F ) x 17 = I X 580
When external torque is removed , only friction creates torque reducing its speed to zero in 120 s so
Angular impulse = change in angular momentum
F x 120 = I X 580
( 46 - F ) x 17 = F x 120
137 F = 46 x 17
F = 5.7 Nm
b )
Putting this value in first equation
5.7 x 120 = I x 580
I = 1.18 kg m²
Answer:
a) τ₁ = 660 N m, b) τ’= 686 N m, c) F = 623.6 N
Explanation:
a) For this exercise let's use the concepts of torque and rotational balance.
For this we set a reference system at the base and assuming that the counterclockwise rotations are positive
where the force F = 600 N, the distance to the axis is x = 1.1 m, the mass of the system m = 70g and the weight is placed at the point of the center of gravity x_{cm} = -1.0 m
The torque at the front is
τ₁ = F x
τ₁ = 600 1.1
τ₁ = 660 N m
b) let's write the rotational equilibrium condition
∑ τ = 0
τ'- W x_{cm} = 0
τ ’= mg x_{cm}
τ’= 70 9.8 1.0
τ’= 686 N m
c) the greatest force Matt can apply
τ’= F x
F = τ’/ x
F = 686 / 1.1
F = 623.6 N