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lord [1]
2 years ago
7

How many half-lives have passed if a sample contains 25% of its original carbon-14?

Physics
1 answer:
ICE Princess25 [194]2 years ago
6 0

Answer: 2 half lives

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A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 H
Ganezh [65]

Answer:

<em> 508Hz</em>

Explanation:

A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 Hz. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - this phenomenon is beat production

frequency is the number of oscillation a wave makes in one seconds.

f1-f2=beats

therefore f1=512Hz

f2=?

beats=4Hz

512Hz-f2=4Hz

f2=512-4

f2=508Hz

the original frequency of the violin is 508Hz

6 0
2 years ago
A box weighing 210 N is pushed up an inclined plane that is 2.0 m long. A force of 140 N is required. If?
N76 [4]
D
Step-by-step explanation:
3 0
3 years ago
Read 2 more answers
"If there were no atmosphere, the tiny molecules of gases and dust particles in the air break up and scatter the light of the su
ELEN [110]

if there was no atmosphere,the tiny molecules of gases and dust particles will gather together and rain down while the sun shine on the particles making the color glow because the colors are combining into various of colors looking like the galaxy.

5 0
2 years ago
A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux
Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

\Phi = EA cos \theta

where

E is the electric field

A is the cross-sectional area

\theta is the angle between the direction of the electric field and the normal to A

The flux is maximum when \theta=0^{\circ}, so we are in this situation and therefore cos \theta =1, so we can write

\Phi = EA

Here we have:

\Phi = 7.50\cdot 10^5 N/m^2 C is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2

And so, we can find the magnitude of the electric field:

E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C

3 0
3 years ago
Question:
KengaRu [80]
Answers:

1A) Al203

1B) SF6

2) Fe203 - iron oxide
3 0
3 years ago
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