All categories

2 years ago

Is velocity ratio of a machine affected by applying oil on it?Explain with reason.

1 answer:

6 0

Factors affecting friction

The intensity of friction depends on following factors: i) The area involved in friction. ii) The pressure applied on the surfaces. Force = Pressure ´ Area Frictional force will increase, if the area of contact will increase or if pressure applied on the surface increased.

Methods to reduce friction

i) Polish the contact surface. ii) Put oil or grease so that it fills in the small gaps of the flat parts. iii) Use ball bearings to reduce area of contact between rotating parts.

Lubrication

Following methods can be used to reduce friction: Oil is either thin or viscous. It depends upon SAE No. of oil. (SAE means Society of Automotive Engineers). If we use very viscous oil, it does not reach all the parts. Very thin oil will flows away easily and gets wasted. Grease is used in such cases. It is generally used around ball-bearing. Normal grease or oil is never used where there is high pressure, high temperature and high speed. Special lubricants are used in such cases. In cold season the oil becomes thick and in hot season it becomes thin. Therefore selection of lubrication also depends on the season. It is always advisable to refer operating manual of the equipment before selecting the lubricant.

You might be interested in

A 26 cm object is 18 cm in front of a plane mirror. A ray of light strikes the object and is reflected off the mirror at a 42-de

matrenka [14]

Answer:

42 degrees, virtual image, same size as the object (26 cm)

Explanation:

The law of reflection states that:

- When a ray of light is incident on a flat surface (such as the plane mirror), the angle of reflection is equal to the angle of incidence

So, since in this case the angle of incidence is 42 degrees, the angle of reflection is also 42 degrees.

Moreover, the image formed by a plane mirror is always:

- Virtual (on the same side as the object)

- The same size as the object

So in this case, since the object's size is 26 cm, the image's size is also 26 cm.

8 0

2 years ago

Glider‌ ‌A‌ ‌of‌ ‌mass‌ ‌0.355‌ ‌kg‌ ‌moves‌ ‌along‌ ‌a‌ ‌frictionless‌ ‌air‌ ‌track‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.095‌ ‌m/s.‌

NemiM [27]

Answer:

vB' = 0.075[m/s]

Explanation:

We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.

Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

$(m_{A}*v_{A})+(m_{B}*v_{B})=(m_{A}*v_{A'})+(m_{B}*v_{B'})$

where:

mA = 0.355 [kg]

vA = 0.095 [m/s] before the collision

mB = 0.710 [kg]

vB = 0.045 [m/s] before the collision

vA' = 0.035 [m/s] after the collision

vB' [m/s] after the collison.

The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.

$(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]$

7 0

2 years ago

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu

Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 $m/s^2$

Explanation:

Mass (m) of the sailboat = 2000 kg

Force acting on the sailboat due to ocean tide is $F_1$ = 3000N

Eastwards means takes place along the positive x direction

Then $F_{1x}$ = 3000N and $F_{1y}$ = 0

Wind Force acting on the Sailboat is $F_2$ = 6000N directed towards the northwest that means at an angle 45 degree above the negative x axis

Then

$F_{2x}$ = -(6000N) cos 45 degree = -4242.6 N

$F_{2y}$ = (6000N) cos 45 degree = 4242.6 N

Hence , the net force acting on the sailboat in x direction is

$F_x = F_{1x}+ F_{2x}$

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is

$F_y = F_{1y}+ F_{2y}$

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

$\sqrt{(1242.6)^2 + (4242.6)^2$

$\sqrt{(1544054.76) + (17999654.8)}$

$\sqrt{(19543709.5)^2}$

4420.8 N

F = ma

$a = \frac{F}{m}$

$a =\frac{4420.8}{ 2000}$

=2.2 $m/s^2$

4 0

2 years ago

A projectile is fired from ground level with a speed of 150 m/s at an angle 30.° above the horizontal on an airless planet where

77julia77 [94]

Answer:

129.9 m/s = 130 m/s to 2 s.f

Explanation:

For projectile motion, the initial velocity combined with the angle of launch is used to obtain the initial horizontal and vertical components of the velocity.

u = initial velocity of the projectile = 150 m/s

uₓ = u cos θ = 150 cos 30° = 129.9 m/s

uᵧ = u sin θ = 150 sin 30° = 75.0 m/s

In the motion of a projectile, the motion can literally be separated into vertical and horizontal components.

The vertical component has to do with the acceleration due to gravity (acting downwards) on the projectile and the vertical component of the velocity (which changes all through the motion of the projectile because of the force of gravity manifested in the form of acceleration due to gravity).

But there is no force acting in the horizontal direction, hence, no acceleration in the horizontal direction for projectile motion. This directly translates to a constant velocity in the horizontal direction all through the flight of the projectile.

Hence, the horizontal component of the velocity of the projectile at t = 4 s is the same as the horizontal component of the initial velocity.

Horizontal component of the Velocity at t = 4s is equal to uₓ = u cos θ = 150 cos 30° = 129.9 m/s = 130 m/s

6 0

2 years ago

Read 2 more answers

A 1.0 kg block is attached to an unstretched

KonstantinChe [14]

Answer:

Change in potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = 1 Nm.

Explanation:

Here, spring constant, k = 50 N/m.

given block comes down eventually 0.2 m below.

here, g = 10 m/s.

let block be at a height h above the ground in figure 1.

⇒In figure 2, potential energy of the block-spring-Earth

system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.

⇒ Change in potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)

= (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.

4 0

2 years ago

Is velocity ratio of a machine affected by applying oil on it?Explain with reason.​

Is velocity ratio of a machine affected by applying oil on it?Explain with reason.