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RSB [31]
3 years ago
11

When lubricating parts that do not have a grease relief port, inject grease until

Engineering
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

C. The boot swells slightly

Explanation:

Friction can be defined as a force that opposes the motion of an object in relation to another.

This ultimately implies that, friction is the opposition to the relative motion of two objects in contact. Therefore, lubrication (lubricants) are used to prevent friction between moving parts.

Generally, grease relief port are created by manufacturers of mechanical parts so as to prevent over-lubrication.

However, when lubricating parts that do not have a grease relief port, inject grease until the boot swells slightly so as to maintain the seal at the ball joints.

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You push a 10 n object 10 m how much work was done on the object​
dezoksy [38]

Answer: the answer is

Work= Force * Distance

Work= 10N * 10m

Work= 100J

Explanation:

5 0
3 years ago
Read 2 more answers
The Hubble space telescope is a bus-sized satellite powered by huge ___ panels
Artemon [7]

Answer:

Solar

Explanation:

The Hubble space telescope is a bus-sized satellite powered by huge solar panels.

4 0
3 years ago
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What is the impedance of a 5μF capacitor at a frequency of 500Hz? What is the impedance of a60mH inductor at this frequency?
choli [55]

Answer:

1) 63.66 ohm

2) 188.49 ohm

Explanation:

Data provided in the question:

Part 1

Capacitance, C = 5μF = 5 × 10⁻⁶ F

Frequency = 500 Hz

Now,

Impedance = \frac{1}{2\times\pi\times f\times C}

or

Impedance = \frac{1}{2\times\pi\times500\times5\times10^{-6}}

or

Impedance = 63.66 ohm

Part 2

Inductance = 60 mH = 60 × 10⁻³ H

Frequency = 500 Hz

Now,

Impedance for an inductor = 2πfL

thus,

Impedance = 2 × π × 500 × 60 × 10⁻³

= 188.49 ohm

4 0
3 years ago
According to the
zysi [14]

Answer:

The part of the system that is considered the resistance force is;

B

Explanation:

The simple machine is a system of pulley  that has two pulleys

The effort, which is the input force at A gives the value of the tension at C and  D which are used to lift the load B

Therefore, we have;

A = C = D

B = C + D = C + C = 2·C

∴ C = B/2

We have;

C = B/2 = A

Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B

The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done

It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force

Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.

3 0
3 years ago
Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa m . It has bee
GaryK [48]

Answer:  Fracture will not occur since Kc (32.2 MPa√m) ∠ KIc (35  MPa√m).

Explanation:

in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.

let us begin,

from the question we have that;

stress = 325 MPa

fracture toughness (KIc) = 35  MPa√m

the max internal crack length = 1.0 m

using the formula;

Y = KIc/σ√(πα)    ---------------(1)

solving for Y we have;

Y =  35 (MPa√m) / 250 (MPa) √(π × 2×10⁻3/2m)

Y = 2.50

so to calculate the fracture roughness;

Kc = Y × σ√(πα)   = 2.5 × 3.25√(π × 1×10⁻³/2) = 32.2 MPa√m

Kc = 32.2 MPa√m

From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35  MPa√m) of the material.

cheers i hope this helps!!!!

8 0
3 years ago
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