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Yuliya22 [10]
3 years ago
13

Care should be exercised when grinding and measuring the following on the Acme threading tool. Which is correct? A. Back rake, t

hread angle, and relief angles B. End flat width, relief angle, and thread angle C. Side rake angle, thread angle, and relief angles D. Side rake, back rake, and relief angles
Engineering
1 answer:
timurjin [86]3 years ago
6 0

Answer:

B). End flat width, relief angle and thread angle

Explanation:

Special care should be taken when grinding and measuring the Acme threading tool because after machining the Acme threading tool, the end flat width, relief angle and the thread angle can change the dimensions of the work piece. So precision measures should be observed while machining as altering these parameters can result in the product with a dimension other than the required dimensions.

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The combustion chamber has different shapes depending on the make and model of the engine. True or false
Alisiya [41]

Answer:

svdsdfdfsdfssdf

Explanation:

fsdsdfsdffsdsfdsdf

5 0
3 years ago
You are hitting a nail with a hammer (mass of hammer =1.8lb) the initial velocity of the hammer is 50 mph (73.33 ft/sec). The ti
Archy [21]

Answer:

The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.

Explanation:

Since we know that the force is the rate at which the momentum of an object changes.

Mathematically \overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}

The momentum of any body is defines as \overrightarrow{p}=mass\times \overrightarrow{v}

In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as

\overrightarrow{F}=\frac{m(\overrightarrow{v_{f}}-\overrightarrow{v_{i}})}{\Delta t}

\overrightarrow{F}=\frac{m(0-(-73.33)}{0.23}=\frac{1.8\times 73.33}{0.23}=573.88Pounds

6 0
3 years ago
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
To revise a monthly budget, changes in which categories might need to be addressed? Check all that apply.
GREYUIT [131]

Income

amount budgeted

expenses

5 0
3 years ago
Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i
IceJOKER [234]

Answer:

Gc(s) = \frac{0.1s + 0.28727}{s}

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

8 0
3 years ago
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