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3 years ago

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A student is studying the ways different elements are similar to one another. Diagrams of atoms from four different elements are

shown below.
Which two atoms are of elements in the same group in the periodic table?

1 answer:

IRISSAK [1]3 years ago

3 0

Answer:atom one and atom four

Explanation:

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can you fill in the blank ____ is the difference between one point of a wave to the same point in the next wave

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The period is the temporal difference between two same points in consecutive waves

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3 years ago

A teacher will never give a student any additional information about a test in a one-on-one meeting because it would not be fair

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False

Explanation:

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3 years ago

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. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg </em>

<em>x1 = -0.75 m </em>

<em>x2 = -0.2 m </em>

<em>g = 9.8 m/s^2 </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

3 0

3 years ago

You hang a heavy ball with a mass of 94 kg from a platinum rod 3.2 m long by 2.2 mm by 3.1 mm. You measure the stretch of the ro

vlada-n [284]

Answer:

$1.67\times 10^{11} N/m^2$

Explanation:

Mass of the ball, $m=94 kg$

Force = weight of the ball

$F=mg = 94kg\times 9.81 m/s^2=922.14N$

For Young's modulus, substitute the values in the formula which is as follows:

$\frac{F}{A}=Y\frac{\Delta L}{L}\\\Rightarrow Y= \frac{FL}{A\Delta L}\\\Rightarrow Y =\frac{922.14 \times 3.2}{2.2\times 10^{-3}\times 3.1\times 10^{-3} \times 0.002588}=1.67\times 10^{11} N/m^2$

4 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

4 years ago