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3 years ago

Find all of those elements on the periodic table

Physics

1 answer:

Vlad [161]3 years ago

6 0

Answer:

hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, argon, potassium, calcium, scandium, titanium, vanadium, chromium, maganese, iron, cobalt, nickel, copper, zinnc, gallium, germanium, aresnic, selenium, bromine, krypton, rubidium, strongtium, yttrium, zirconium, niobium, molybdenum, technetium, ruthenium, rhodium, palladium, silver, cadmium, indium, tin, atimony, tellurium, iodine, xenon, cesium, barium, lanthanum, cerium, praseodyumium, neodymim, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, luteium, hafnium, tantalum, tungsten, rhenium, osmium, irdium, platinum, gold, mercury, thallium, lead bismuth, polnium, astatine, radon, rancium, radium, actinum, thorium, protactinium

Explanation:

You might be interested in

Does heat demagnetize?

Nastasia [14]

Answer:

Yes

Explanation:

Heat affects the magnets because it confuses and misaligns the magnetic domains, causing magnetism to decrease

3 0

3 years ago

1. What is energy? What can an object with energy do?

kvv77 [185]

Answer:

Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed.

3 0

3 years ago

Avery is experimenting with a simple circuit. She measures the current in the circuit three different time

sammy [17]

The current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.

Ohm's Law states that the voltage across an electric conductor is directly proportional to the current(I) passing through it provided the resistant is constant.

So;

V ∝ I

V = IR

where

R = constant

The objective of this question want us to determine: How did the current change for each test provided that Avery uses a 1.5-volt battery, then she uses a 3-volt battery and lastly she uses a 9-volt battery, given that the resistance is constant through out the whole process.

In the first experiment;

1.5 = IR

In the second experiment;

3 = IR

In the third experiment;

9 = IR

Therefore, we can conclude that the current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.

Learn more about Ohm's Law here:

brainly.com/question/14296509

5 0

3 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

8 0

3 years ago

The air in a kitchen has a mass of 60.0kg and a specific heat of 1505J/(kg°C).

BARSIC [14]

Your answer is 632,100J which is Choice D

8 0

4 years ago

Read 2 more answers