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Oliga [24]
2 years ago
10

A diffusion couple composed of two silver– gold alloys is formed; these alloys have compositions of 98 wt% Ag–2 wt% Au and 95 wt

% Ag– 5 wt% Au. Determine the time this diffusion couple must be heated at 750°C (1023 K) in order for the composition to be 2.5 wt% Au at the 50 μm position into the 2 wt% Au side of the diffusion couple. Preexponential and activation energy values for Au diffusion in Ag are 8.5 × 10−5 m2/s and 202,100 J/mol, respectively.
Physics
1 answer:
jeyben [28]2 years ago
4 0

Answer:

for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)

It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.

we have D= D0exp( -Qd/RT)

=(8.5×105m2/s)exp(-202,100/8.31×1023)

= 4.03 ×10-15m2/s

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Alyssa is carrying a water balloon while running down a field at a speed of 14 m/s. She tosses the water balloon forward toward
Luda [366]
From Alyssa's point of view, the water balloon is at first at rest and then gets thrown with a velocity of 23m/s. Therefore the balloon will have a speed of 23m/s for Alyssa.

At the same time, Naya is watching, and she sees the balloon at the beginning moving at a speed of 14m/s along with Alyssa, and then pushed forward of other 23m/s. Therefore, from her point of view, the balloon will have a speed of 14+23 = 37m/s.

Hence, the correct answer is <span>D) The speed of the balloon is 23 m/s for Alyssa and 37 m/s for Naya. </span>
4 0
2 years ago
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For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




7 0
2 years ago
Read 2 more answers
Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.
dimulka [17.4K]

Answer:

The distance of separation is d = 0.092 \ m

Explanation:

The mass of the each ball is  m= 10 g  =  0.01 \ kg

 The negative charge on each ball is q_1 =q_2=q =  1 \mu C  =  1 *10^{-6} \ C

Now we are told that the lower ball is  restrained from moving this implies that the net force acting on it is  zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

          F =  \frac{kq_1 * q_2}{d}

=>       m* g  =  \frac{kq_1 * q_2}{d}

here k the the coulomb's  constant with a value  k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

So  

      0.01 * 9.8  =  \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}

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5 0
2 years ago
what is the acceleration of a bowling ball that starts at rest and moves 300m down the gutter in 22.4 sec
exis [7]
<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>
7 0
2 years ago
A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?
andreev551 [17]

<span>Hello!
 
We have the following data:
</span>
Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

P =  \frac{E}{T}

Solving:

P = \frac{E}{T}

P = \frac{E}{T} \to T =  \frac{E}{P}

T =  \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}

\boxed{T = 3000\:seconds}

How many minutes can it run for? (<span>Let's convert in minutes)
</span>
1 minute --------- 60 seconds
y minute --------- 3000 seconds

\frac{1}{y} = \frac{60}{3000}

<span>Product of extremes equals product of means
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60*y = 1*3000

60y = 3000

y =  \frac{3000}{60}

\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark


I hope this helps! =)
<span>

</span>
7 0
2 years ago
Read 2 more answers
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