Answer:
The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.
Explanation:
Given that,
Mass of the meter stick, m = 0.3 kg
Center of mass is located at its 45 cm mark.
We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :

or

So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.
1. clavicle = collarbone
2. vertebrae = backbone
3. scapula = shoulder blade
4. femur = thigh
5. humerus = upper arm
6. patella = kneecap
7. cranium = skull
8. tibia = lower leg
9. radius/ulna = forearm
10. phalanges = fingers/toes
Answer:
The stress is calculated as 
Solution:
As per the question:
Length of the wire, l = 75.2 cm = 0.752 m
Diameter of the circular cross-section, d = 0.560 mm = 
Mass of the weight attached, m = 25.2 kg
Elongation in the wire, 
Now,
The stress in the wire is given by:
(1)
Now,
Force is due to the weight of the attached weight:
F = mg = 
Cross sectional Area, A = 
Using these values in eqn (1):
Answer:
<h2>b) 4230 J
</h2>
Explanation:
Step one:
given data
extension= 40cm
Spring constant K= 52.9N/cm
Step two:
Required
the Kinetic Energy KE
the expression to find the kinetic energy is
KE= 1/2ke^2
substituting our data we have
KE= 1/2*52.9*40^2
KE=0.5*52.9*1600
KE= 42320Joules
<u>The answer is b) 4230 J
</u>
Answer:
14 rev
Explanation:
= initial angular velocity = 2.5 revs⁻¹
= final angular velocity = 0.8 revs⁻¹
= Angular acceleration = - 0.2 revs⁻²
= Angular displacement
Using the equation

So the number of revolutions are 14