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Talja [164]
3 years ago
15

The fundamental frequency of a musical note is 330Hz. Which harmonic is this?

Physics
1 answer:
allsm [11]3 years ago
6 0

Answer:

it is a type of even harmonic

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Why might measuring changes in speed or direction be important?
aliya0001 [1]
It’s important because if you were let’s say training to see how far you go in 10sec you need to record your data to see how much you are improving in that skill or something
5 0
3 years ago
Which way does a river flow??
spin [16.1K]

Downwards - from uphill towards the lowlands and eventually into the sea.

8 0
3 years ago
Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla
svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

6 0
3 years ago
Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an
lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

5 0
3 years ago
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