One is the best choice for an answer.
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.
Answer:
a)The direction the frictional force will acts is in the positive x direction.
Explanation:
a)The direction the frictional force will acts is in the positive x direction
b)in the horizontal direction, the total force F(total) is equal to 4times the frictional force in the wheel.
F(total)=4f
''f'' is taken as the frictional force.
c)4times the normal force on each wheel minus the acceleration equals zero i.e 4N(wheel)-a=0
=4N(wheel)-mg=0
d) torque is the force that tends to bend rotation
ζ=rf
but acceleration=4×frictional force
cross multiply
f=ζ/r
f=ma/4
ma/4=ζ/r
a=4ζ/r
ΔH<span> for the reaction = [sum of heat of formation for products] - [sum of heat of formation for reactants]
-6534 = [(12)(-393.5) + (6)(-285.8)] - (2X)
-6534 = [(-4722) + (-1714.8)] - 2X
-6534 = -6436.8 - 2X
-6534 + 6436.8 = -2X
-</span>97.2 = -2X<span>
then, divide each side by two to cancel out the numerical coefficient.
x = +48.6 kJ, which is the heat of formation for benzene.</span>
