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Jobisdone [24]
2 years ago
5

How many 10" diameter circles can be cut from a semicircular shape that has a 20"

Engineering
2 answers:
Iteru [2.4K]2 years ago
5 0

Answer:

<h2> 1</h2>

  • <em>The</em><em> </em><em>diameter</em><em> </em><em>measurement</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>semi circle</em><em> </em><em>having</em><em> </em><em>a</em><em> </em><em>measure</em><em> </em><em>of</em><em> </em><em>1</em><em>0</em><em>"</em><em> </em><em>diameter</em><em> </em><em>,</em><em> </em><em>2</em><em>0</em><em>"</em><em> </em><em>diameter</em><em> </em><em>and</em><em> </em><em>a</em><em> </em><em>length</em><em> </em><em>of</em><em> </em><em>2</em><em>5</em><em>.</em>
N76 [4]2 years ago
3 0

9514 1404 393

Answer:

  1

Explanation:

Only one such circle can be drawn. The diameter of the 10" circle will be a radius of the semicircle. In order for the 10" circle to be wholly contained, the flat side of the semicircle must be tangent to the 10" circle. There is only one position in the figure where that can happen. (see attached).

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DE QUE MANERA LA ALEGRIA NOS AYUDA A CONSEGIR AMIGOS <br> ≤→ω←≥
madreJ [45]

Answer: Nos hace más valientes y nos empuja a hacerlo. >_<

Explanation:

6 0
3 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

Answer:

62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = 10^5\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

1\ \text{mile}=1609.34\ \text{m}

\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

Dislocation density = 10^{10}\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

3 0
3 years ago
Compared to 15 mph on a dry road, about how much longer will it take for
Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
Read 2 more answers
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
"A communication between two devices is over the maximum limit of an ethernet frame size. The Transmission Control Protocol (TCP
Sliva [168]

COmmunication Devices

Explanation:

  • TCP/IP, or the Transmission Control Protocol/Internet Protocol, is a suite of communication protocols used to interconnect network devices on the internet. TCP/IP can also be used as a communications protocol in a private network (an intranet or an extranet).

  • The sequence number in a header is used to keep track of which segment out of many this particular segment might be. The next field, the acknowledgment number, is a lot like the sequence number. The acknowledgment number is the number of the next expected segment.

  • The 32 bit sequence number field defines the number assigned to the first byte of data contained in this segment. TCP is a stream transport protocol. To ensure connectivity, each byte to be transmitted is numbered.
3 0
3 years ago
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