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Verdich [7]
3 years ago
8

One gram of Strontium-90 has an activity of 5.3 terabecquerels (TBq), what will be the activity of 1 microgram?

Engineering
1 answer:
noname [10]3 years ago
8 0

1 micro gram of Strontium-90 has an activity of

0.0000053 terabecquerels (TBq),

Explanation:

Given information denotes that .,one gram of Strontium-90 has an activity of 5.3 terabecquerels (TBq)

the activity of 1 micro gram is

1 gram = 1,000,000 micro gram has activities of 5.3 terabecquerels

therefore 1 micro gram has the activity of (5.3 ÷  1,000,000 = 0.0000053 )

= (5.3 ÷  1,000,000 = 0.0000053 )

Hence ., 1 micro gram of Strontium-90 has an activity of

0.0000053 terabecquerels (TBq),

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I think it’s B, digital

Explanation:

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What are the 3 dimensions that used in isometric sketches?
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Answer:

The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D

Explanation:

An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.

5 0
3 years ago
Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a
Arada [10]

Answer:

a) h_c = 0.1599 W/m^2-K

b) H_{loss} = 5.02 W

c) T_s = 302 K

d) \dot{Q} = 25.125 W

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, T_a = 25^0 C = 298 K

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600}  = 1.25 kg/sec

Rate of heat transfer,

\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W

a) To calculate the convection coefficient relationship for heat transfer by convection:

\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, k_c = 0.8

\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K

b) Heat loss from the pipe to the environment:

H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W

d) The required fan control power is 25.125 W as calculated earlier above

5 0
3 years ago
Pls list up to five key lessons and knowledge areas that you have acquired in this course about operations management. How do yo
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Explanation:

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3 0
3 years ago
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Zepler [3.9K]

Answer:

591.3

Explanation:

99.19 + (1.85 × 266) = 591.29

rounded = 591.3

4 0
4 years ago
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