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strojnjashka [21]
2 years ago
11

If the frequency of a FM wave is 8.85 × 107 hertz, what is the period of the FM wave?

Physics
1 answer:
patriot [66]2 years ago
8 0

Answer:

1.1299 x 10^-8   second    

Explanation:

Period = 1 / f = 1 / (8.85 * 10^7) = 1.1299 x 01^-8  sec

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A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?
fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as V_1, the relation with the final volume as

V_2 = V_1 +0.163\% V_1

V_2 = V_1 +0.00163V_1

V_2 = 1.00163V_1

Initial temperature = 21.1\°C

Let T be the temperature after expanding by the formula of volume expansion

we have,

V_2 = V_1 (1+\gamma \Delta t)

Where \gamma is the volume coefficient of copper 5.1*10^{-5}/C

1.00163V_1 = V_1(1+\gamma(T-21.1\°))

1.00163 = 1+5.1*10^{-5}(T-21.1\°)

0.00163 = 0.000051T-0.0010761

T = 53.0608\°C

Therefore the temperature is 53.06°C

7 0
3 years ago
Two equal positive charges are held in place at a fixed distance. If you put a third positive charge midway between these two ch
eimsori [14]

Answer:

False

Explanation:

8 0
3 years ago
Un paracaidista desciende desde 6 000 m de altura. Si la masa, con su equipo, es de 65 kg, ¿cuánto valdrá su energía potencial e
iris [78.8K]

Explanation:

PRIMERO ENCUENTRAS EL PESO DEL PARACAIDISTA

F_{peso} = 65 kg(9.80 m/s^{2}) = 637 N

CON LA FÓRMULA DE LA ENERGÍA POTENCIAL

U = 637 N(6000 m - 3500 m) = 1592500 J

3 0
3 years ago
Inside most ball-point pens is a small spring that compresses as the pen is pressed against the paper. If a force of 0.1 N compr
AnnZ [28]

Answer:

20 N/m

Explanation:

From the question,

The ball-point pen obays hook's law.

From hook's law,

F = ke............................ Equation 1

Where F = Force, k = spring constant, e = compression.

Make k the subject of the equation

k = F/e........................ Equation 2

Given: F = 0.1 N, e = 0.005 m.

Substitute these values into equation 2

k = 0.1/0.005

k = 20 N/m.

Hence the spring constant of the tiny spring is 20 N/m

8 0
3 years ago
What is the first step of thermonuclear fusion within the Sun to form helium-4?
hodyreva [135]

Great Question! I happened to be a physics nerd!

Answer:

C. Two hydrogen nuclei, each with only one proton, fuse to form deuterium, a form of hydrogen with one proton.

MAKE SURE TO SEE EXPLANATION!

Explanation:

In the core of the Sun, or any other main sequence star, there is no single fusion process. Instead, complex sequences of processes occur to make helium nuclei from hydrogen nuclei (i.e. protons). The proton-proton chain provides for the majority of energy generation in stars with masses less than that of the Sun.  One difficulty in creating a helium nucleus (two protons and two neutrons) is that there are only protons to begin with. Some protons must be turned into neutrons in some way. The first step is to combine two protons to form a deuterium nucleus (also known as a deuteron). That's a hefty hydrogen nucleus with one proton and one neutron. Such a proton-proton contact is highly unlikely, and it has never been detected in a laboratory. Fortunately, the Sun's core is incredibly hot and dense, with an incredible number of protons packed inside. Even a low likelihood event will occur every now and again. Along with each deuteron, a positron (an "anti-electron") and a neutrino are created. Because the Sun's core is plasma, there are a lot of free electrons, thus the positron doesn't live long until it and an electron collide and annihilate, resulting in gamma radiation. The deuteron then interacts with a proton to form a helium 3 nucleus. That is a high-probability interaction, and it occurs swiftly. Two helium 3 nuclei join in the third phase to generate a helium 4 ("regular" helium) nucleus and a proton. Branch I of the proton-proton (p-p) chain is responsible for this. Another stage is required because reactions between helium 3 and helium 4 nuclei are possible. There are two conceivable reactions (named Branch II and Branch III), and I'll save you the gory details. It gets much more complicated since theoretical calculations indicate that a reaction between a helium 3 nucleus and a proton is feasible — Branch IV. This reaction has an incredibly low likelihood of occurring, far lower than the Branch I reaction, thus it must be exceedingly rare. The Carbon-Nitrogen-Oxygen (CNO) Cycle is another method for reducing hydrogen to helium. It does not generate much energy in the Sun, but it is the principal energy generation mechanism in larger stars.

8 0
2 years ago
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