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Alexandra [31]
4 years ago
9

As microwave light travels through a liquid, it moves at a speed of 2.2 x 108 m/s. If the frequency of this light wave is 1.1 x

108 Hz, what is the wavelength of this microwave?
Physics
1 answer:
Alexandra [31]4 years ago
7 0

λ = 2 m.

The easiest way to solve this problem is using the equation of frecuency of a wave f = v/λ, where v is the velocity of the wave, and λ is the wavelength.

To calculate the wavelength of a microwave light travels through a liquid, it moves at a speed of 2.2 x 10⁸ m/s. If the frecuency of the light wave is 1.1 x 10⁸ Hz, we have to clear λ from the equation f = v/λ:

f = v/λ -------> λ = v/f

λ = 2.2 x 10⁸ m/s / 1.1 x 10⁸ Hz

λ = 2 m (wavelength of the microwave)

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What is the production​ function? The production function is the relationship between
Ludmilka [50]

Answer:

The Production Function is the relationship between the quantity or variables of output and the different inputs quantities used in the production process.

Explanation:

Generally in quality, production or any business, we need to know what is the output with respect to different input variables. Suppose if an output is about bottle produce per hour, then we look at the inputs such as temperature, raw material/inventory, power supply, manpower, etc which are needed to produce bottle. any changes, for example in temperature could change the production rate.

Furthermore, a graph is made to see the production function.

7 0
3 years ago
Gamma rays may be used to kill pathogens in ground beef. One irradiation facility uses a 60Co source that has an activity of 1.0
kirill115 [55]

Answer:

Explanation:

C_i=3.7\times 10^{16} \,decays/sec

Energy of gamma rays due to ore decay is

(1.17+1.33)MeV=2.50MeV

Energy of gamma ray produced in seconds is

(2.5\times 3.7\times 10^{16})MeV

Activity

1 \times 10^6c

Total energy of gamma ray produced in second is

(2.5 \times 3.7 \times 10^{16} \times 10^6)MeV

E-energy of gamma ray produced in an hour is

(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV

Gamma ray energy absorbed by meat = 30% (in 1 hour) of E is 0.3E

Dose required to kill pathogen = 4000Gy=4000J/kg

The kilogram of meat that can be produced per hour is

\frac{0.3E/hr}{4000J/kg}\\\\E=(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV\\\\=333\times 10^{18}MeV\\\\eV=1.6\times 10^{-19}\\\\E=53.3J

The meat that can be produced =\frac{0.3\times 53.3}{4000}=3.996\times 10^{-3}

4 0
4 years ago
A 3.0 kilogram object is thrown upward with an initial speed of 10 meters per second. what maximum height will the object reach?
tino4ka555 [31]
Time= v*sintheta/g = 10*1/9.8= 1.02s therefore h=1/2*9.8*(1.02)^2= 5.10m
3 0
3 years ago
A 0.850 kg block is attached to a spring with spring constant 18 N/m . While the block is sitting at rest, a student hits it wit
Sveta_85 [38]

The block's speed at the point where x=0.25A is v = 31.95 cm/s.

<h3>What is Spring constant?</h3>

The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the bigger the spring constant.

question is incomplete, this is the remaining statement

What is the amplitude of the subsequent oscillations? And What is the block's speed at the point where x=0.25A?

x = Asin(wt)

v = Aw coswt

at t = 0

w = sqrt(k/m)

v = Aw

A = v/w

A = 7.17 cm

part b )

E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2

mv^2 + k(1/4A)^2 = 1/2kA^2

mv^2 + kA^2/16 = kA^2

mv^2 = kA^2 - kA^2/16

mv^2 = 15kA^2/16

v^2 = 15/16 * (k/m) * A^2

v^2 = 15/16 *w^2A^2

v = sqrt(15/16) * wA

v = 31.95 cm/s

to learn more about spring constant go to -

brainly.com/question/23885190

#SPJ4

8 0
2 years ago
all bearings are made by lebng spherical drops of molten metal fall inside a tall tower – called a shot tower – and solidify as
Wewaii [24]

Answer:

Part b)

h = 78.5 m

Part c)

v = 39.24 m/s

Explanation:

Part b)

If ball need t = 0 to t = 4 s then height of the tower is the total displacement of the ball in t = 4 s interval

here if ball start from rest

then its displacement is given as

\Delta y = \frac{1}{2}gt^2

\Delta y = \frac{1}{2}(9.81)(4^2)

\Delta y = 78.5 m

Part c)

Speed of the bearing at the end of the motion of the ball

v_f = v_i + at

v_f = 0 + (9.81)(4)

v_f = 39.24 m/s

7 0
3 years ago
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