A. electrons<span> and </span>neutrons<span> B. </span>electrons<span> and </span>protons<span> C. </span>protons<span> and </span>neutrons<span> D. all particles are attracted to each other. According to atomic theory, </span>electrons<span> are usually found: A. in the </span>atomic nucleus<span> B. outside the nucleus, yet very near it because they are attracted to the </span>protons<span>.</span>
The object's final velocity, given the data is 10.5 rad/s
<h3>What is acceleration? </h3>
This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
- a is the acceleration
- v is the final velocity
- u is the initial velocity
- t is the time
<h3>How to determine the final velocity</h3>
The following data were obtained from the question
- Initial velocity (u) = 1.5 rad/s
- Acceleration (a) = 0.75 rad/s²
- Time (t) = 12 s
- Final velocity (v) = ?
The final velocity can be obtained as follow:
a = (v – u) / t
0.75 = (v – 1.5) / 12
Cross multiply
v – 1.5 = 0.75 × 12
v – 1.5 = 9
Collect like terms
v = 9 + 1.5
v = 10.5 rad/s
Thus, the final velocity of the object is 10.5 rad/s
Learn more about acceleration:
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Answer:
= 391.67 Hz
Explanation:
The sound of lowest frequency which is produced by a vibrating sting is called its fundamental frequency (
).
The For a vibrating string, the fundamental frequency (
) can be determined by:
= 
Where v is the speed of waves of the string, and L is the length of the string.
L = 42.0 cm = 0.42 m
v = 329 m/s
= 
= 
= 391.6667 Hz
The fundamental frequency of the string is 391.67 Hz.
Answer:
self-contained, short-term light-source
Explanation:
Answer:
the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Explanation:
Given the data in the question;
wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m
Index of refraction; n = 1.35
Now, the thinnest thickness of the soap film can be determined from the following expression;
= ( λ / 4n )
so we simply substitute in our given values;
= ( 463 × 10⁻⁹ m ) / 4(1.35)
= ( 463 × 10⁻⁹ m ) / 5.4
= ( 463 × 10⁻⁹ m ) / 4(1.35)
= 8.574 × 10⁻⁸ m
= 85.74 × 10⁻⁹ m
= 85.74 nm
Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm