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mojhsa [17]
3 years ago
9

Hard woods and tight curves should be cut using what blade speeds.

Engineering
1 answer:
erica [24]3 years ago
4 0

Answer:

slower

Explanation:

When it comes to using scroll saws, it is important to consider the <u>type of wood</u> in order to choose the correct speed. The blades of the saw tends to break easily, thus it is essential to choose a <em>"slower speed"</em> when shaping hard woods. Such speed is also used for<em> "tight curves."</em> Furthermore, it is important not to force the wood to the blade because it will also cause breakage.

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Suzanne Brett wants to borrow $55,000 from the bank. The interest rate is 6.5% and the term is for 5 years.
Otrada [13]

Answer: $14575

$55000

6.5%

5 years

Total Payment Amount: $72875

Yearly payment :$72875/5= $14575

8 0
2 years ago
What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-
Hitman42 [59]

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}

Where \sigma_{m} is the magnitude of he maximum stress at the tip of the crack, \sigma_{p} is the magnitude of the tensile stress, l_{c} is 1/2 the length of the internal crack, and r_{c} is the radius of curvature of the crack.

We have:

r_{c}=1.9*10^{-4} [mm]

l_{c}=3.8*10^{-2} [mm]

\sigma_{c}=140 [MPa]

We replace:

\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}

We get:

\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]

5 0
3 years ago
A source current of 10 mA is supplied to a parallel circuit consisting of the following resistors three resistors, a 2200 a 500
marysya [2.9K]
It’s is 2000 because it is 20000
7 0
3 years ago
A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude
Fofino [41]

Answer:

u = 260.22m/s

S_{max} = 1141.07ft

Explanation:

Given

S_0 = 89.6ft --- Initial altitude

S_{16.5} = 0ft -- Altitude after 16.5 seconds

a = -g = -32.2ft/s^2 --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

S = ut + \frac{1}{2}at^2

The final altitude after 16.5 seconds is represented as:

S_{16.5} = S_0 + ut + \frac{1}{2}at^2

Substitute the following values:

S_0 = 89.6ft       S_{16.5} = 0ft     a = -g = -32.2ft/s^2    and t = 16.5

So, we have:

0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2

0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45

0 = 89.6 + 16.5u-  4383.225

Collect Like Terms

16.5u = -89.6 +4383.225

16.5u = 4293.625

Make u the subject

u = \frac{4293.625}{16.5}

u = 260.21969697

u = 260.22m/s

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

v=u  + at

At the maximum height:

v =0 --- The final velocity

u = 260.22m/s

a = -g = -32.2ft/s^2

So, we have:

0 = 260.22 - 32.2t

Collect Like Terms

32.2t = 260.22

Make t the subject

t = \frac{260.22}{ 32.2}

t = 8.08s

The maximum height is then calculated as:

S_{max} = S_0 + ut + \frac{1}{2}at^2

This gives:

S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2

S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22

S_{max} = 89.6 + 260.22 * 8.08 - 1051.11

S_{max} = 1141.0676

S_{max} = 1141.07ft

Hence, the maximum height is 1141.07ft

8 0
3 years ago
Identify 3 distractions for young drivers and explain how you plan to minimize these distractions
Veseljchak [2.6K]

Answer:

Phone, Eating, People in the car

Explanation:

your phone should always be off while driving and put away. Eating much like texting takes your attention away from the road so you should wait till you're parked or home to eat. People thrashing around or talking very loudly can be a terrible distraction. Ask them nicely to quiet down so you can focus on the road.

4 0
3 years ago
Read 2 more answers
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