Just some random stufff

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

noname [10]

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),

8 0

3 years ago

Can I get an answer to this question please

crimeas [40]

Answer:

(i) 12 V in series with 18 Ω.

(ii) 0.4 A; 1.92 W

(iii) 1,152 J

(iv) 18Ω — maximum power transfer theorem

Explanation:

As seen by the load, the equivalent source impedance is ...

10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

(36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

The load current is ...

(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

(600 s)(1.92 J/s) = 1,152 J

__

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

7 0

2 years ago

PLEASE HELP!! Its easy!!!

Rina8888 [55]

Answer:

C is tire

F is cassette

D is hub

4 0

3 years ago

Read 2 more answers

⦁ An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and

Anon25 [30]

Answer:

<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</em></u>

Download docx

8 0

4 years ago

120 litres of water is discharge from container in 25 seconds. Find the rate of discharge in cumecs.if the discharge took place

notsponge [240]

<h2>Answer:</h2>

Rate of discharge in cumecs: <u>0.0048m³/s</u>.

Velocity flow: <u>24m/s</u>.

<h2>Explanation:</h2>

<h3>1. Find the rate of discharge in cumecs.</h3>

a. Convert from litres to m³.
120L*1000= 120000mL

120000mL=120000cm³

120000cm³/100³=0.12 m³.

b. Rate of discharge.

<em>If 0.12 m³ where discharged in 25 seconds, the rate of discharge is:</em>

0.12m³/25s = 0.0048m³/s.

<em />

<h3>2. Find the velocity flow.</h3>

Let's refer to the fluid mechanics equation that relates volume flow, area and velocity. This is the formula:

$\frac{dV}{dt}=Av$ ; where the expression $\frac{dV}{dt}$ is the volume flow rate (in m³/s); A is the cross-sectional area of the pipe (in m²), and v is the velocity flow (in m/s).

a. Solve the equation for v.

$\frac{dV}{dt}=Av\\ \\(\frac{dV}{dt})/A=v\\ \\v=(\frac{dV}{dt})/A$

b. Calculate the cross-sectional area of the pipe.

<em>The cross-sectional area of the pipe is a circle. Hence, the formula of this area is:</em>

$A=\pi r^{2}$

<em>We'll have to convert the diameter to meters, because the formula for flow velocity needs the area in m². Let's go ahead and do that.</em>

<em /> $50mm/1000=0.05m$ .

<em>We were given the diameter, and the formula uses the radius, but the radius is just half of the diameter, therefore, we can substitute in toe formula like this:</em>

$A=\pi (\frac{0.05}{2} )^{2}=0.0020m^{2}$

c. Substitute in the new expression for velocity flow and calculate.

$v=(\frac{dV}{dt})/A\\ \\v=(\frac{0.048m^{3} }{1s})/(0.0020m^{2} )\\\\ v= 24m/s$

8 0

2 years ago