A string is rolled around a cylinder (m = 3.9 kg) as shown in the figure below.

Question

A string is rolled around a cylinder (m = 3.9 kg) as shown in the figure below. A person pulls on the string, causing the cylinder to roll without slipping along the table. If there is a tension of 34 N in the string, what is the acceleration of the cylinder? A board of length 1.7 m is attached to a floor with a hinge at one end as shown in the figure below. The board is initially at rest and makes an angle of 55 degree with the floor. If the board is then released, what is its angular acceleration? ANSWER IS IN RAD/S^2 For the system below, m1 = 12 kg, m2 = 21 kg and mpulley = 23 kg, if the coefficient of friction between crate 1 and the table is ?K = 0.14, what is the acceleration? Calculate the moment of inertia of a CD, including the effect of the hole. For a CD of radius 3.0 cm, estimate the percentage change in the moment of inertia due to a hole of radius 7 mm. (Neglect the difference in mass due to the hole.) A solid cube of mass 35 kg and edge length 0.55 m rests on a horizontal floor as shown below. A person then pushes on the upper edge of the cube with a horizontal force of magnitude F. At what value of F will the cube start to tip? Assume the frictional force from the floor is large enough to prevent the cube from sliding. Consider the flagpole in the figure below. If the flagpole has a mass of 16 kg and length 10 m and the angle the cable makes with the pole is Phi = 24 degree , what are the magnitude and direction of the force exerted by the hinge (at point P) on the flagpole? Assume the mass of the pole is distributed uniformly. A tree grows at an angle of 50 degree to the ground as shown in the figure below. If the tree is 29 m from its base to its top and has a mass of 450kg, what is the magnitude of the torque on the tree due to the force of gravity? Take the base of the tree as the pivot point. (Assume the tree is a long clinder of uniform density. The answer reveals one reason trees need roots.)

Explanation / Answer

mercury(II) oxide molar mass= 216.59 g mol?1

90 gm of mercury(II) oxid=90/216.59=0.4155mole

2HgO ==> 2Hg (liquid) + O2 (gas)

2mole HgO produces 1mole O2 gas

hence,

0.4155 mole HgO produces 0.2078mole O2

at 25oC and 1 atm 1mole O2 occupied volume=22.4L

0.2078 mole O2 volume=22.4*0.2078L=4.654L (ans)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.