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yanalaym [24]
1 year ago
14

Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to

structure, properties, and processing. Which of the following is the best example of a material property?
a. Polycrystalline
b. Sintering Annealing
c. Melting Temperature
d. Casting
Physics
1 answer:
Vera_Pavlovna [14]1 year ago
5 0

Answer:

c. Melting Temperature

Explanation:

  • Polycrystalline this is a solid material like common salts, ceramics, rocks, and ice that are composed of many crystalline of varying size and orientation.
  • Sintering Annealing is the use of heat to remove internal stresses from certain materials
  • Melting temperature also known as melting point,  is the temperature at which a substance changes from solid to liquid state. Every solid material has a melting point. It is a material property.
  • Casting this is a manufacturing process in which a liquid material is usually poured into a mold, which contains a hollow cavity of the desired shape, and then allowed to solidify.

Therefore, the correct option is 'C' Melting Temperature

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All of these are examples of pseudopsychology EXCEPT:
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psychoanalysis

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A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p
DedPeter [7]

Answer:

a) v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s}  =3.2051 \frac{m}{s}

b) v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s}  =5.2083 \frac{m}{s}

c) a=\frac{v_{2}-v_{1}  }{t_{2}-t_{1}  } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }

Explanation:

<em><u>The knowable variables are </u></em>

d_{t}=25m

t_{1}=1.3 s

t_{2}=3.9 s

t_{3}=5.5 s

Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is \frac{25}{3} m</u>

a) v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s}  =3.2051 \frac{m}{s}

b) v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s}  =5.2083 \frac{m}{s}

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) a=\frac{v_{2}-v_{1}  }{t_{2}-t_{1}  } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }

6 0
1 year ago
8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

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6 0
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timurjin [86]

Answer:

F₁ = F₂ = F₃ = 0 N

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Arrow 3 mass = 90 g   speed = 9 m/s

Horizontal Force:- F₁ , F₂ and F₃

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If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.

We know,

According to newton's second law

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If Acceleration is equal to zero

Then Force is also equal to zero.

Hence, F₁ = F₂ = F₃ = 0 N

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