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Savatey [412]
3 years ago
14

Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C.

The carriage and half-nuts D. A1l of the above
Engineering
1 answer:
Maksim231197 [3]3 years ago
8 0

Answer:

(d) all of the above

Explanation:

before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial

so option (D) will be correct because we need lubricate in all the given parts  

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Batter boards (or battre boards, Sometimes mispronounced as "battle boads") are temporary frames, set beyond the corners of a planned foundation at precise elevations. These batter boards are then used to hold layout lines (construction twine) to indicate the limits (edges and corners) of the foundation.

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2 years ago
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Merchants get commission from selling products that are consigned to them true or false​
Marta_Voda [28]

Answer:

true

Explanation:

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3 years ago
Calculate the magnitude of the velocity and the θ angular direction of the block and the bullet together when the 50 g bullet mo
almond37 [142]

Answer:

Magnitude of the velocity = 16.82 m/s

Angular direction, θ = 52.41°

Explanation:

As given ,

mass of bullet, m₁= 50g = 0.05 kg

speed of bullet , u₁ = 600 m/s

mass of the block , m₂ = 4 kg

speed of the block before collision , u₂ = 12 m/s

direction , θ = 30°

Now,

Assume that the combined velocity of bullet and block after collision = v

and the direction = θ

Now, from the conservation of momentum in x - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ

where v = final velocity after collision

u₁ = initial velocity of bullet before collision = 0

m₁ = mass of the bullet before collision = 0.05 kg

u₂  = velocity of block before collision = 12 cos(30° )

m₂ = mass of block before collision

m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ

⇒ 0 + 4(6√3) = 4.05 vₓ

⇒24√3 = 4.05 vₓ

⇒vₓ = 10.26 m/s

Now, from the conservation of momentum in y - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) v_{y}

where v = final velocity after collision

u₁= initial velocity of bullet before collision = 600

m₁ = mass of the bullet before collision = 0.05 kg

u₂  = velocity of block before collision = 12 sin(30° )

m₂= mass of block before collision

m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 ) v_{y}

⇒ 30 + 4(6) = 4.05 v_{y}

⇒30 +24 = 4.05 v_{y}

⇒54 = 4.05 v_{y}

⇒v_{y} = 13.33 m/s

Now, the magnitude of the velocity = √vₓ² + v_{y}² = √(10.26)² + (13.33)²

                                                           = √105.26 + 177.68

                                                           = √282.95 = 16.82

The angular direction, θ =  tan^{-1}(\frac{v_{y} }{v_{x} }) =  tan^{-1}(\frac{13.33}{10.26}) = tan^{-1}(1.299) = 52.41°

8 0
3 years ago
A 8.0-m-diameter parachute of a new design is to be used to transport a load from flight altitude to the ground with an average
S_A_V [24]

Answer:

drag coefficient for the parachute is Cd = 1.84177

Explanation:

given data

diameter = 8.0 m

average vertical speed = 3 m/s

total weight of load = 200 N

to find out

drag coefficient for the parachute

solution

we will apply here drag coefficient formula that is express as here  

drag coefficient Cd = \frac{2w}{\rho v^2A}    ......................1

here w is total load and A is area and v is speed

and here ρ = 1.22 kg/cu

put here value we get  

Cd = \frac{2*4*200}{1.229*5^2*\pi * 3^2}  

Cd = 1.84177

drag coefficient for the parachute is Cd = 1.84177

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Explanation:

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