Two horizontal forces are exerted on a large crate. The first force is 317 N to the right. The second force is 173 N to the left

Response:

2.5kN.m

Details:

Torque relates directly to the pitch diameter

= Ta/Tb= Da/Db

For 120/Tb= 0.25/0.5

This gives Tb= 2.469kN.m, roughly 2.5kN.m

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

• 52.92% metal: 52.92 g M
• 47.80% oxygen: 47.80 g O

 By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

Given 52.92 g of metal in the sample, the molar mass of the metal is:

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

Answer:

a = 18.28 ft/s²

Explanation:

the values provided are:

duration of force application, t= 10 s

Work done = 10 Btu

mass of the object = 15 lb

acceleration, a =? ft/s²

1 Btu = 778.15 ft.lbf

thus, 10 Btu = 7781.5 ft.lbf

m = 0.466 slug

So,

the work is equivalent to the change in kinetic energy

 

The acceleration of the object is therefore

 

 

         a = 18.28 ft/s²

the constant acceleration of the object is calculated to be 18.28 ft/s²

Factors influencing friction

The magnitude of friction is contingent on the following elements: i) The surface area in contact. ii) The applied pressure on the surfaces. Force is determined by Pressure multiplied by Area; thus, if the contact area increases or if the pressure applied rises, the frictional force will also escalate.

Methods for reducing friction

i) Smooth the contact surface. ii) Apply oil or grease to fill small gaps in flat surfaces. iii) Use ball bearings to minimize contact area among rotating components.

Lubrication

To minimize friction, various methods may be employed: Oil can be either thin or viscous, which depends on its SAE number (SAE indicating Society of Automotive Engineers). Highly viscous oils may not reach all components effectively. In contrast, very thin oils may drain away quickly, resulting in wastage. Grease is preferable in such situations, particularly around ball-bearings. Regular grease or oil should not be utilized under high speed, high pressure, and high temperature conditions—specialized lubricants are required then. The consistency of oil varies with temperature; it thickens in the cold and thins in the heat. Therefore, the choice of lubricant should be seasonally appropriate, and it's always wise to consult the equipment's operating manual prior to making a selection.[[TAG_11]]

The force can be determined using the equation F (force) = mass * acceleration. The unit of measurement, N (Newton), is equivalent to kilogram-meter/seconds2.
Thus, F= 1300 kg * 1.07 m/s2 = 1391 N.
The resultant value is 1391 N.