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Doss [256]
3 years ago
12

An outdoor Wi-Fi unit for a picnic area has a 110 mW output and a range of about 38 m. What output power (in mW) would reduce it

s range to 18 m for use with the same devices as before? Assume there are no obstacles in the way and that microwaves into the ground are simply absorbed.
Physics
1 answer:
valkas [14]3 years ago
8 0

Answer:

The output power is 24.68 mW.

Explanation:

Given that,

Power = 110 mW

Range = 38 m

Reduced range = 18 m

We need to calculate the power

Using formula of intensity

I=\dfrac{P}{A}

I=\dfrac{P}{\pi r^2}

As intensity is constant

P\propto r^2

So, \dfrac{P_{1}}{P_{2}}=\dfrac{r_{1}^2}{r_{2}^{2}}

P_{2}=\dfrac{r_{2}^2}{r_{1}^{2}}\timesP_{1}

Put the value into the formula

P_{2}=\dfrac{18^2}{38^2}\times110\times10^{-3}

P_{2}=24.68\times10^{-3}\ W

P_{2}=24.68\ mW

Hence, The output power is 24.68 mW.

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The upward force exerted on an object falling through air is _____.
stira [4]

(C) Air Resistance

<u>Explanation:</u>

When an object falls through air, air resistance acts on it in upward direction. When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.

6 0
3 years ago
Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.
Fudgin [204]

Answer:

g'_h=1.096\times 10^{-5}\ m.s^{-2}

Explanation:

We know that the gravity on the surface of the moon is,

  • g'=\frac{g}{6}
  • g'=1.63\ m.s^{-2}

<u>Gravity at a height h above the surface of the moon will be given as:</u>

g'_h=\frac{G.m}{(r+h)^2} ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

  • G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}
  • m=7.35\times 10^{22}\ kg
  • r=1.74\times 10^6\ m
  • h=384.4\times 10^6\ m is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}

g'_h=1.096\times 10^{-5}\ m.s^{-2} is the gravity on the moon the earth-surface.

4 0
3 years ago
Why electrons flow in a wire when connected to a battery?
vesna_86 [32]

Answer:

The metal atoms in the wire can't move, but their outer electrons can. The force pushes those electrons and they move to further parts of the wire, trying to reach the other end. As the electrons move away, new electrons flow into the wire through the battery to take their place.

Explanation:

6 0
3 years ago
What is E of a hydrogen atom in the 3p state?
notsponge [240]

Answer:

E=-1.51 eV.

L=\hbar\sqrt{2}

Explanation:

The nth level energy of a hydrogen atom is defined by the formula,

E_{n}=-\frac{13.6}{n^{2} }

Given in the question, the hydrogen atom is in the 3p state.

Then energy of n=3 state is,

E_{n}=-\frac{13.6}{(3)^{2} }\\E_{n}=-1.51eV

Therefore, energy of the hydrogen atom in the 3p state is -1.51 eV.

Now, the value of L can be calculated as,

L=\hbar\sqrt{l(l+1)}

For 3p state, l=1

L=\hbar\sqrt{1(1+1)}\\L=\hbar\sqrt{2}

Therefore, the value of L of a hydrogen atom in 3p state is L=\hbar\sqrt{2}.

4 0
3 years ago
A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total
Aloiza [94]

Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

6 0
2 years ago
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