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2 years ago

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2. Other igneous rock forms from lava that cools quickly on Earth’s surface. Classify the rock as either intrusive or extrusive,

and identify another word used to describe this type of rock.

1 answer:

Darya [45]2 years ago

5 0

Answer:

extrusive,

Explanation:

lava exlodes outwards, making it extrusive and not intrusive.

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A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq

Semmy [17]

Answer:

I=9.6×e^{-8} A

Explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

=4.11×e^-6×I

we have that

(flux)'= -U

so differentiating flux we get

so the inducted emf in the loop.

U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

so, I=2.9×e^{-6}÷30

I=9.6×e^{-8} A

6 0

3 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

Amanda [17]

Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603 225

0.203 368

0.162 442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

$Bending Stress = \frac{1}{(Mirror Radius)^{n}}$

At mirror radius 1 = 0.603 mm Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

$\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}$

$\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}$

$0.6114 = (0.3366)^{n_1}$

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

$\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}$

$\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}$

$0.8326 = (0.7980)^{n_2}$

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

$\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}$

$\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}$

$0.5090 = (0.2687)^{n_3}$

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

$n = \frac{n_1 + n_2 + n_3}{3}$

$n = \frac{0.452 +0.821 + 0.514}{3}$

n = 0.596

Hence to get bending stress x at mirror radius 0.796

$\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}$

$\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}$

$\frac{Stress x}{225} = 0.8475$

stress x = 191 MPa

3 0

3 years ago

Machine movement can be divided into what two main categories?

pishuonlain [190]

Answer:

motion and power

Explanation:

8 0

3 years ago

Read 2 more answers

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and

k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

$z1=$ $\frac{\left(150-137\right)}{27.7}$ $=0.4693$

$z2=\frac{\left(201-137\right)}{27.7}=2.3105$

$P(0.4693$

B.)

$z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289$

$\\P(2.9309$

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

3 0

3 years ago

Read 2 more answers

Radio Frequency IDentification (RFID) tags and readers are a category of low-end wireless devices that people may not recognize

Fittoniya [83]

Answer:

See explaination.

Explanation:

Radio Frequency Identification (RFID) tags and readers uses the electromagnetic waves to identify and track the attached objects.

A tag is attached to the object which is to be identified or tracked, and reader is used to read the response and send the acknowledgement. Therefore, RFID tags and readers are used in many industries, passports, transportations and pet identification etc.

i.

RFID technology is used in smartcards, implants for pets, passports and library books to identify and track the persons, objects and pets etc.

Hence, we can say that option (i) is true.

ii.

Electronic Product Code (EPC) is a small code stored in the RFID tag. The code stored in the memory is 96 bits which are used to identify the organization which manages the data, unique number to identify the product and a number to identify the particular tag and etc.

EPC is a unique identification number, it can read and be written by the RFID reader. It is used in supply chains instead of a bar code even though expansive.

Hence, option (ii) is true.

iii.

Tags are used to identify and track the objects. It doesn’t belong to base stations and access points as a Wi-Fi networks.

Therefore, option (iii) is false.

iv.

The EPC Generation 2 RFID tag is used to improve the security by enabling the authentication features. It is not the similar way of data transmission in the other wireless situations.

Therefore, option (iv) is false.

Finally, the options (i) and (ii) are TRUE, while (iii) and (iv) are FALSE.

8 0

3 years ago