Answer:
I do believe her actions were justified.
Explanation:
Due to the school charging extra fee from her father who makes a modest amount as a teacher. There was sum of money involved that could change how he lived and her.
I do not believe her actions where justified
She had a lot going for her. She could have skipped the hardship of helping grace and pass. She could have easily have gotten a good job with a degree and paid back all the debts owed. Alot of troubles could have been avoided just by doing her own thing.
Answer:
2.1 rad(anticlockwise).
Explanation:
So, we are given the following data or parameters or information in the question above:
=> "The torsional stiffness of the spring support is k = 50 N m/rad. "
=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"
=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"
Hence;
G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.
Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).
==> 0.15/20 (V - w) + θ = 0.
==> 0.15/20 (V - w ) = -θ.
Where V = k = 50 N m/rad
w = 183.3 θ.
Therefore, w + Vθ = 500 Nm.
==> 183.3 + 50 θ = 500 Nm.
= 6.3
Anticlockwise,
θ = 2.1 rad.
Answer:
A. -5488J
B. 273.8J
C. 372.44N
Explanation:
Given:
m = 40kg
h = 14 m
v= 3.7 m/s
Part(a)
The change in the potential energy of the bear Earth system during the slide
AU = -mgh = -40(9.8) (14) = -5488 J
Part(b)
The kinetic energy of the bear just before hitting the ground is
Ks 1/2 mV^2= (40)(3.7)2 = 547.6 /2 = 273.8J
Part(c)
The change in the thermal energy of the system due to friction is
AEth = fxh=-(AK +AU) = 5488– 273.8 = 5214.2 J
The average frictional force that acts on the sliding bear is
F = Eth / 14= 5214.2/14 =372.44N
Answer:
hope its helpful to uh...
