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To which group/family does each of these belong? A. Sulfur _________ B. Sodium _________ C. Argon _________ D. Silicon _________

ArbitrLikvidat [17]

Answer:

A. Sulfur _________ group 16 chalcogens

B. Sodium _________ group 1 alkali metals

C. Argon _________ group 18 noble gases

D. Silicon _________ group 14 carbon family

E. Chlorine _________ group 17 halogens

F. Phosphorus_________ group 15 pnicogens

7 0

1 year ago

In a nuclear reactor, the control rods are: A.made of graphite B.used to keep the reactor cool C.used to absorb free neutrons D.

algol13

I think its E. the control rods are used to control <span>the fission rate of uranium and plutonium. Hope this helps!! </span>

4 0

3 years ago

Read 2 more answers

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

What type of radioactive decay is shown in this equation?

svp [43]

There is no <span>radioactive decay</span>

6 0

2 years ago

An elastic rope is tied securely to a wall. An upward-displaced pulse is introduced into the rope. The pulse travels through the

Natasha2012 [34]

Answer:

Inverted (displaced downwards)

Explanation:

The pulse becomes INVERTED upon reflecting off the boundary with the wall. That is, an upward-displaced pulse will reflect off the end and return with a downward displacement. This inversion behavior will always be observed when the end of the medium is fixed, like this wall in this instance. This INVERSION BEHAVIOR can also be observed when the medium is connected to another more heavy or more dense medium. And in this case, when the pulse reaches the end of the medium, a portion of the pulse will reflect off the end and return with an inverted displacement. The heavier medium acts like a fixed end to cause the pulse to be inverted.

Summary: a pulse reaching the end of a medium becomes inverted whenever it either:

i. reflects off a fixed end,

ii. is moving in a less dense medium and reflects off a more dense medium.

3 0

2 years ago