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trasher [3.6K]
2 years ago
14

Examples of reciprocating motion in daily life

Engineering
1 answer:
bonufazy [111]2 years ago
8 0

Answer:

Examples of reciprocating motion in daily life are;

1) The needles of a sewing machine

2) Electric powered reciprocating saw blade

3) The motion of a manual tire pump

Explanation:

A reciprocating motion is a motion that consists of motion of a part in an upward and downwards (\updownarrow) or in a backward and forward (↔) direction repetitively

Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump

In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.

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-1 1/6 divided by 2 1/3
pantera1 [17]

Answer:

-\frac{1}{2}

Explanation:

-1\frac{1}{6}:2\frac{1}{3}=\\  \\=-\frac{7}{6}:\frac{7}{3}\\  \\=-\frac{7}{6}*\frac{3}{7}\\  \\=-\frac{1}{2}

4 0
3 years ago
What happens to battery when it produces current to the system
olasank [31]

Answer:

Electricity, as you probably already know, is the flow of electrons through a conductive path like a wire. This path is called a circuit. ... The chemical reactions in the battery causes a build up of electrons at the anode. This results in an electrical difference between the anode and the cathode

3 0
3 years ago
At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing
Inessa [10]

Answer:

vB = - 0.176 m/s   (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒  (OA) = Sin 35°*(AB)

⇒   (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒  (OB) = Cos 35°*(AB)

⇒   (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒  (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒  vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒  vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒  vB = - 0.176 m/s   (↓-)

The pic can show the question.

7 0
3 years ago
Read 2 more answers
A home electrical system is joined to the electric company's system at the junction of the
aleksandrvk [35]

That would be B, I hope this helps!

5 0
2 years ago
A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f
jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

8 0
3 years ago
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