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2 years ago

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Faça uma redação sobre '' A luta das mulheres brasileiras pelos seus direitos ''

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Kazeer [188]2 years ago

5 0

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a 1000 khz am radio station broadcasts with a power of 20 kw. how many photons does the transmitter antenna emit each second

Gemiola [76]

1000 khz am radio station broadcasts with a power of 20 kw number of photon emitted per second is 30.16 x 10^30 photon/s.

The frequency of the radio station is:
f
=
1000
k
H
z
=
1
×
10^6Hz
The transmit power is: P = 20kW = 20 X 10^3 W

The transmit power is: h = 6.63 x 10 ^-34 m^2.kg/s

The number of photon emitted per second = N = P/hf = <u>30.16 x 10^30 </u>photon/s.

1000 khz am radio station broadcasts with a photon of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.

Learn more about photon on:
brainly.com/question/20912241

#SPJ4

6 0

6 months ago

A dock worker loading crates on a ship finds that a 27 kg crate, initially at rest on a horizontal surface, requires a 80 N hori

Aleksandr-060686 [28]

Answer:0.302

Explanation:

Given

mass of crate m=27 kg

Force required to set crate in motion is 80 N

Once the crate is set in motion 56 N is require to move it with constant velocity

i.e. 80 N is the amount of force needed to just overcome static friction and 56 is the kinetic friction force

thus

$f_s(static\ friction)=\mu \cdot N$

where $\mu$ is the coefficient of static friction and N is Normal reaction

$N=mg$

$f_s=\mu mg$

$\mu mg=80$

$\mu =\frac{80}{27\times 9.8}$

$\mu =0.302$

7 0

3 years ago

In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an

natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

$A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}$

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

$=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa$

σ=5517.25 Pa

Strain in x direction

ε=σ/E

$=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}$

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0

2 years ago

Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o

zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

$F= ma$

$a=\dfrac{F}{m}$

Put the value of F

$a=\dfrac{kq^2}{mr^2}$

Put the value into the formula

$414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}$

$q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}$

$q^2=1.84\times10^{-14}$

$q=0.135\times10^{-6}\ C$

$q=0.135\ \mu C$

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0

2 years ago

Read 2 more answers

What is the split atom called?

son4ous [18]

If you're talking about the <em>splitting</em> of an atom, the process is called Nuclear Fission.

3 0

2 years ago

Read 2 more answers