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andrew11 [14]
3 years ago
10

If the frequencies of two component waves are 24 Hz and 20 Hz, they should produce _______ beats per second.

Physics
2 answers:
horrorfan [7]3 years ago
8 0
This can be answered using the beat frequency formula, which is simply the difference between 2 frequencies.

Let: <span>fᵇ = beat frequency
</span>f₁ = first frequency
f₂ = second frequency

fᵇ = |f₁ - f₂|

substituting the values:
fᵇ = |24Hz - 20Hz|
fᵇ = 4Hz

The unit Hz also means beats per second, therefore:
<span>fᵇ = 4 beats per second
</span>
Therefore, the answer is C. 4
mrs_skeptik [129]3 years ago
3 0

C. 4 beats per second is correct

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Dmitriy789 [7]

Answer:

3.1 × 10^- 7 m and 2.1 × 10^-7 m

Explanation:

First we must convert each value of energy to Joules by multiplying its value by 1.6 ×10^-19. After that, we can now obtain the wavelength from E= hc/λ

Where;

h= planks constant

c= speed of light

λ= wavelength of light

For 6.0ev;

E= 6.0 × 1.6 ×10^-19

E= 9.6 × 10^-19 J

From

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/9.6 × 10^-19

λ= 2.1 × 10^-7 m

For 4.0 eV

4.0 × 1.6 × 10^-19 = 6.4 × 10^-19 J

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/6.4 × 10^-19

λ= 3.1 × 10^- 7 m

3 0
3 years ago
The equation for free fall at the surface of a celestial body in outer space​ (s in​ meters, t in​ seconds) is sequals10.04tsqua
Sindrei [870]

Answer:

1.42 s

Explanation:

The equation for free fall of an object starting from rest is generally written as

s=\frac{1}{2}at^2

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t is the time

On this celestial body, the equation is

s=10.04 t^2

this means that

\frac{1}{2}g = 10.04

so the acceleration of gravity on the body is

g=2\cdot 10.04 = 20.08 m/s^2

The velocity of an object in free fall starting from rest is given by

v=gt

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

v = 28.6 m/s

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t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s

3 0
3 years ago
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

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W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

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The workdone by the gas on the bullet is 12,279.75J

3 0
3 years ago
Which is not a factor that affects the pressure of a gas in a closed container?
maksim [4K]

it is size of of particles because it does not matter about the size in a closed container

7 0
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Ronch [10]

Answer:

a.  μ_{95%} = 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Explanation:

a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:

μ_{95%} = x_±(t*s)/sqrt(n)

where:

μ_{95%} = = is the 95% confidence interval estimate

x_ = mean of the sample = 3

s = standard deviation of the sample = 5.8

n = size of the sample = 41

t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021

substituting all the variable, we have:

μ_{95%} = 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.

4 0
3 years ago
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