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Alexeev081 [22]
3 years ago
10

the average time required for radioactive elements, whereby they decay to half their original mass is called the "average life"

of the element
Physics
2 answers:
Setler [38]3 years ago
7 0
False.

<span>The time required for radioactive elements, whereby they decay to half their original mass is called the "Half life" of the element

Not the average life.
</span>
VikaD [51]3 years ago
4 0
I'm not sure what your question is. But, the half life is the amount of time required for half the material to decay. For U238 this is 4.5 billion years, whilst for Fr-223 (Francium) its about 22 minutes. To calculate the time for something to decay you need to use the equation:
Mass (after time t) = Mass (initial) * (0.5)^(time/half life)
Hope this helps
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1. The current in a wire is 0.72A. Calculate the charge that passes through the wire in; a. 4 s b. 60 s c. 180 s d. 7 s e. 0.5 s
Vlad [161]
Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t

Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C

Case II
----------
When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C

Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C

Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C

Case V
----------
When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
6 0
3 years ago
Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is c_p=1.0\ Btu/lbm.F, and the steam properties as, A-4E:

h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0
3 years ago
Can someone give me tips for science camp!! Please
Kamila [148]
Try to have a calm morning before camp. A good night’s sleep and a good breakfast. Make sure to be cautious, follow all the rules for certain areas ( some maybe restricted ). Take lots of photos doing wacky stuff! Learn but have fun learning
5 0
2 years ago
Read 2 more answers
Based on the situation above, choose the CORRECT type of error.
tatiyna

Answer:

what's your question I can't understand

5 0
2 years ago
6. The momentum of a 30.0 g bird with a speed of 12 m.s-1 is 0.36 kg.m.s-1. What will be its momentum 12s later if a constant .0
Keith_Richards [23]

Answer:

Explanation:

initial momentum = .36 kg.m.s⁻¹

negative impulse = force x time = .02 x 12 = .24 kg.m.s⁻¹

final momentum - initial momentum = impulse

final momentum = initial momentum + impulse

= .36 - .24

= .12 kg.m.s⁻¹

7 0
2 years ago
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