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3 years ago

What is a chipping hammer used for? State three things.

Engineering

2 answers:

aleksley [76]3 years ago

7 0

Answer:

i hope this helps.

Explanation:

they are used for breaking concrete, can be positioned to break vertical and overhead surfaces, allows precisely chip away only specific areas.

avanturin [10]3 years ago

7 0

Chipping hammers are used for a multitude of reasons some are to break concrete, break welds in automotive quality processes, clean large grey iron castings of flash, clean foundry furnaces, chip cast iron pipe. Anywhere you are trying to remove or clean something which requires heavy duty removal of materials.

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What are the functions of each computer program

Ludmilka [50]

A function is a block of organized, reusable code that is used to perform a single, related action. Functions provide better modularity for your application and a high degree of code reusing. ... Different programming languages name them differently, for example, functions, methods, sub-routines, procedures, etc.

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3 years ago

A screw is a simple machine that can be described as an inclined plane wrapped around a cylinder.

Phoenix [80]

Answer:

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3 years ago

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mamaluj [8]

Answer:

Explanation:

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3 years ago

Read 2 more answers

The flexural strength or MOR of a ceramic is 310 MPa. A block of the ceramic, which is 20 mm wide, 15 mm high, and 300 mm long,

Alja [10]

Answer:

$F=6200\ \text{N}\\$

Explanation:

In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:

In this problem the flexural strength is defined with the following formula:

$\sigma=\cfrac{3FL}{2bd^2}$

where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.

The force is then defined as:

$F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}$

3 0

3 years ago

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago