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rosijanka [135]
1 year ago
6

new generation cordless phones use a 9.00x10^2 MHz frequency and can be operated up to 60.0 m from their base. how many waveleng

ths of the electromagnetic waves can fit between your ear and a base 60.0 m away?
Physics
1 answer:
Musya8 [376]1 year ago
3 0
Λ = 3*10^8 / 9*10^8 = 1/3 m
no. of wavelengths = 60/(1/3) = 180
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What is the concentration of molecular oxygen (O2) in mol/L on a June day in Toronto when atmospheric pressure is 1.0 atm and th
saveliy_v [14]

Answer:

The concentration of mole evil at oxygen on that day is 0.00858 mol/L

Explanation:

Here, we want to calculate the concentration of molecular oxygen

The pressure on that day is 1.0 atm

Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm

Now let’s calculate the concentration;

From Ideal gas law;

PV = nRT

This can be written as;

P/RT = n/V

The term n/V refers to concentration;

Let’s make substitutions now;

P = pressure = 0.21 atm

R = molar gas constant = 0.0821 L•atm/mol•k

T = temperature = 25 = 25 + 273.15 = 298.15 K

Substituting these values, we have;

n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L

6 0
1 year ago
The amount of pressure required to move a 6800 lb force with a 6" d piston is ___ psi.
Katena32 [7]
The pressure needed in PSI = Pounds of force needed divided by the cylinder Area
The Cylinder rod Area is 21.19  sq inches
Thus, the pressure= 6800/21.19
                              = 320.91 PSI

7 0
1 year ago
Which of the following statements cannot be supported by Kepler's laws of planetary motion?
gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio \frac{a^3}{T^2} (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>

Is not supported by any Kepler's law.

8 0
1 year ago
3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg
Anni [7]

Answer:

\mathbf{1.51\times10^{-15}N}

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

q_{paper}=-4.1\times10^{-15}C

Now the distance from the charge is

r=1cm=0.01m

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}

\Rightarrow W=mg

=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}

=\mathbf{1.51\times10^{-15}N}

6 0
1 year ago
The PVT behavior of a certain gas is described by the equation of state: P(V − b) = RT where b is a constant. If in addition CV
alexdok [17]

Answer:

shown in the attachment

Explanation:

The detailed step by step and necessary mathematical application is as shown in the attachment.

6 0
2 years ago
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