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3 years ago

Is NaCH a element?Why or why not?

Physics

2 answers:

sattari [20]3 years ago

8 0

Answer:

Explanation:

I don't think it is, but I do know that NaCI is an element. Hope this helps!

Arisa [49]3 years ago

8 0

Answer:

No, NaCH is not an element.

Explanation:

NaCH is a compound of several elements not just one.

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Where were scientific advancements being made during the dark ages

Leto [7]

Answer:

The period saw major technological advances, including the adoption of gunpowder, the invention of vertical windmills, spectacles, mechanical clocks, and greatly improved water mills, building techniques (Gothic architecture, medieval castles), and agriculture in general (three-field crop rotation).

Explanation:

Hope this helps :)

7 0

3 years ago

Read 2 more answers

You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just

Zepler [3.9K]

Answer:

The angular magnification is $M = 2.808$

Explanation:

From the question we are told

The focal length is $f = 8.9cm$

The near point is $n_p = 25.0cm$

The angular magnification is mathematically represented as

$M = \frac{n_p}{f}$

Substituting values

$M = \frac{25}{8.9}$

$= 2.808$

4 0

3 years ago

What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat

Amiraneli [1.4K]

Answer:633.8 KJ

Explanation:

Given

mass of water $\left ( m\right )=250gm$

Initial temperature $\left ( T_i\right )=20^{\circ}C$

Final temperature $\left ( T_f\right )=100^{\circ}C$

Specific heat of water $\left ( c \right )$ =4190 J/kg-k

heat of vaporization $\left ( L\right )=22.6\times 10^5 J/kg$

Heat required for process $\left ( Q\right )$ =heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc $\left ( T_f-T_i\right )+mL$

Q= $0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5$

Q= $\left ( 0.838+5.5\right )\times 10^5$

Q= $6.338\times 10^5J=633.8 KJ$

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3 years ago

A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped f

dem82 [27]

Answer:

Part a)

$Loss = 3.6 J$

Part b)

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

Explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

$U = mgH$

Energy Loss = 0.35 mgH[/tex]

$Loss = 0.35(0.70)(9.81)(1.5)$

$Loss = 3.6 J$

Part b)

Energy of the ball after first bounce

$U_1 = 0.65 mgH$

energy of ball after 2nd Bounce

$U_2 = 0.65(0.65 mgH)$

energy of the ball after 3rd bounce

$U_3 = (0.65)(0.65^2)mgH$

$U_3 = 0.65^3(0.70)(9.81)(1.5)$

$U_3 = 2.83 J$

Now we will have energy loss in fourth bounce given as

$Loss = 0.35 U_3$

$Loss = 0.35(2.83)$

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

7 0

3 years ago

Calculate the highest frequency X-rays produced by 8.00 · 104 eV electrons. _____ Hz

lys-0071 [83]

When an electron stops, it emits a photon with energy equal to the kinetic energy lost by the electron:
$E=\Delta K$
The energy of the photon is $E=hf$ where $h=6.63 \cdot 10^{-34} Js$ is the Planck constant and f is the frequency. Therefore, the maximum frequency of the emitted photon occurs when the loss of kinetic energy is maximum.

The maximum loss of kinetic energy of the electron occurs when the electron stops completely, so it loses all its energy:
$\Delta K = 8\cdot 10^4 eV$
Keeping in mind that $1 eV=1.6 \cdot 10^{-19}J$ , we have
$\Delta K = 8 \cdot 10^4 eV \cdot 1.6 \cdot 10^{-19}J/eV = 1.28 \cdot 10^{-14}J$

And so, this corresponds to the energy of the emitted photon, E. Therefore, we can find the maximum frequency of the emitted photon:
$f= \frac{E}{h} = \frac{1.28 \cdot 10^{-14}J}{6.63 \cdot 10^{-34}Js}=1.93 \cdot 10^{19}Hz$

5 0

3 years ago