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3 years ago

Before finishing and installing a shelved cabinet you just constructed, you need to check the

Engineering

1 answer:

Greeley [361]3 years ago

7 0

Answer:

Carpenter's square

Explanation:

The most common hand tool used to measure or set angles with its application extending to setting angles of roofs and rafters. Another name of a Carpenter's square is a framing square.

Other hand tools that are used to measure angles are;

The combination square that allows a user to set both 90° and 45° angles
A Bevel that allows users to set any angle they like.
A Protractor that resembles a bevel but its marks are marked in an arc.
An electromagnetic angle finder which gives a reading according to the measure of the arms adjusted by the user.

You might be interested in

Which one of the following is a list of devices from least efficient to most efficient

Sergeu [11.5K]

Answer:

(A)-Light-Bulb, Automobile Engine, Power Plant, Electric Motor

Explanation:

To get the least efficient to the most efficient the chart below explains how to the get it

Incandescent light bulb-5%

Florescent light bulb-20%

Electric motor-90%

Power plant-35%

Automobile engine-25%

Home oil furnace-80%

4 0

3 years ago

A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m

natima [27]

Answer:

a. 0.4544 N

b. $5.112 \times 10^{-5 M}$

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

$H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O$

$NaOH\ Mass = Normality \times equivalent\ weight \times\ volume$

$= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL$

= 0.27264 g

$NaOH\ mass = \frac{mass}{molecular\ weight}$

$= \frac{0.27264\ g}{40g/mol}$

= 0.006816 mol

Now

Moles of $H_2SO_4$ needed is

$= \frac{0.006816}{2}$

= 0.003408 mol

$Mass\ of\ H_2SO_4 = moles \times molecular\ weight$

$= 0.003408\ mol \times 98g/mol$

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

$= \frac{acid\ mass}{equivalent\ weight \times volume}$

$= \frac{0.333984 g}{49 \times 0.015}$

= 0.4544 N

b. And, the acid solution molarity is

$= \frac{moles}{Volume}$

$= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}$

= 0.00005112

= $5.112 \times 10^{-5 M}$

We simply applied the above formulas

4 0

3 years ago

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun

Alex17521 [72]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19, 24, 13 , 29, 19, 21, 14

It is given that,

∑x = 299

∑y = 167

∑ $x^{2}$ = 11887

∑ $y^{2}$ = 3773

We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52

∑x = 299

Yes, it is equal to the given value. Hence, verified.

2. ∑y = 167

Let's verify it:

∑y = 30 + 19 + 24 + 13 + 29 + 19 + 21 + 14

∑y = 169

No, it is not equal to the given value.

3. ∑ $x^{2}$ = 11887

Let's verify it:

For this to find, first we need to square all the value of x individually and then add them together to verify.

∑ $x^{2}$ = $27^{2}$ + $44^{2}$ + $32^{2}$ + $47^{2}$ + $23^{2}$ + $40^{2}$ + $34^{2}$ + $52^{2}$

∑ $x^{2}$ = 11,887

Yes, it is equal to the given value. Hence, verified.

4. ∑ $y^{2}$ = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑ $y^{2}$ = $30^{2}$ + $19^{2}$ + $24^{2}$ + $13^{2}$ + $29^{2}$ + $19^{2}$ + $21^{2}$ + $14^{2}$

∑ $y^{2}$ = 3,845

No, it is not equal to the given value.

4 0

2 years ago

A prototype boat is 30 meters long and is designed to cruise at 9 m/s. Its drag is to be simulated by a 0.5-meter-long model pul

Ghella [55]

Answer:

a) 1.16 m/s

b) 1/216000

c) (√15)/6480000

Explanation:

The parameters given are;

Length of boat prototype, lp = 30 m

Speed of boat prototype = 9 m/s

Length of boat model, lm= 0.5 m

a) lm/lp = 0.5/30 = 1/60 = ∝

(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)

vm = 9 × (1/60)^(1/2) = 1.16 m/s

b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;

Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³

Fm/Fp = (1/60)³ = 1/216000

c) The ratio of the model to prototype power pm/p $_p$ = (Fm/Fp) × (vm/vp) = ∝³×√∝

The ratio of the model to prototype power pm/p $_p$ = √(1/60) × (1/60)³

pm/p $_p$ = √(1/60) × (1/60)³ = (√15)/6480000

6 0

3 years ago

Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a

nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0

3 years ago