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mrs_skeptik [129]
3 years ago
6

Pls help with my physics test?!​

Physics
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer:

hold on bro I got it

d I think is the answer

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Carbon is allowed to diffuse through a steel plate 11 mm thick. The concentrations of carbon at the two faces are 0.88 and 0.41
Serhud [2]

Answer:

The temperature is 2584.5 K

Explanation:

Given:

Activation energy Q = 80000 \frac{J}{mol}

Preexponential D= 6.2 \times 10^{-7} \frac{m^{2} }{s}

Diffusion flux J = 6.4 \times 10^{-10} \frac{kg}{m^{2} s}

Thickness of plate \Delta x =  11 \times 10^{-3} m

Concentration of carbon at two faces \Delta C = (0.88 - 0.41 ) = 0.47 \frac{kg}{m^{3} }

From the formula of temperature in terms of diffusion flux,

  T = (\frac{Q}{R} ) \frac{1}{\ln (\frac{D\Delta C}{J\Delta x} )}

Where R = 8.314 \frac{J}{mol.K} ( gas constant )

Put the values and find the temperature,

  T = (\frac{80000}{8.314} ) \frac{1}{\ln (\frac{6.2 \times 10^{-7} \times 0.47 }{6.4 \times 10^{-10}\times 11 \times 10^{-3} } )}

  T = 2584.5 K

Therefore, the temperature is 2584.5 K

8 0
3 years ago
The acceleration due to gravity on the surface of Venus is 8.83 m/s2. An object with a mass of 5.23 kg has what weight on Venus?
Marat540 [252]
Weight = m times g = 5.23 times 8.83 = 46.18 N
6 0
3 years ago
A car is driving at 99 km/h, calculate the distance it travels in 70 minutes.
Lapatulllka [165]

Answer:

The distance the car travels is 115500 m in S.I units

Explanation:

Distance d = vt where v = speed of the car and t = time taken to travel

Now v = 99 km/h. We now convert it to S.I units. So

v = 99 km/h = 99 × 1000 m/(1 × 3600 s)

v = 99000 m/3600 s

v = 27.5 m/s

The speed of the car is 27.5 m/s in S.I units

We now convert the time t = 70 minutes to seconds by multiplying it by 60.

So, t = 70 min = 70 × 60 s = 4200 s

The time taken to travel is 4200 s in S.I units

Now the distance, d = vt

d = 27.5 m/s × 4200 s

d = 115500 m

So, the distance the car travels is 115500 m in S.I units

8 0
3 years ago
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp
IRINA_888 [86]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?

Answer:

601000 N

Explanation:

Given that :

Acceleration due to gravity at lunar outpost = 1.6m/s²

Supported Weight of supplies = 1 * 10^5 N

Acceleration due to gravity on the earth surface = 9.8m/s²

Maximum weight of supplies as measured on EARTH :

Ratio of earth gravity to lunar post gravity:

(Earth gravity / Lunar post gravity) ;

(9.8 / 1.63) = 6.01

Hence, maximum weight of supplies as measured on EARTH should be :

6.01 * (1 × 10^5)

6.01 × 10^5

= 601000 N

3 0
3 years ago
What is the weight of a dust particle
Alexandra [31]

Answer:

The mass of a dust particle is 7.53 x 10^-10 kg. (Hope it helped.)

Explanation:

3 0
3 years ago
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