Answer:
805.48N/m
Explanation:
According to Hookes law
F = Ke
F is the force = mg
F = 2.4×9.8 = 23.52N
e is the extension = 2.92cm = 0.0292m
Force constant K = F/e
K = 23.52/0.0292
K = 805.48N/m
Hence the force constant of the spring is 805.48N/m
(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².
(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².
(1.c) The relative intensity of the sound as heard by the listener is 103 dB.
(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.
(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.
<h3>
Surface area being vibrated</h3>
The surface area being vibrated by the time the sound reaches the listener is calculated as follows;
A = 4πr²
A = 4π x (20)²
A = 5,026.55 m²
<h3>Intensity of the sound</h3>
The intensity of the sound is calculated as follows;
I = P/A
I = (100) / (5,026.55)
I = 0.02 W/m²
<h3>Relative intensity of the sound</h3>
<h3>Speed of sound at the given temperature</h3>
<h3>Frequency of the sound</h3>
The frequency of the sound heard is determined by applying Doppler effect.
where;
- -v₀ is velocity of the observer moving away from the source
- -vs is the velocity of the source moving towards the observer
- fs is the source frequency
- fo is the observed frequency
- v is speed of sound
Learn more about intensity of sound here: brainly.com/question/17062836
Answer:
<u>In an ionic bond , an element will have to lose or gain electrons.</u>
Explanation:
- Ionic bond, also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.
- Such a bond forms when the valence (outermost) electrons of one atom are transferred permanently to another atom.
- <em>The atom that loses the electrons becomes a positively charged ion (cation), while the one that gains them becomes a negatively charged ion (anion).</em>
∴
- <em>The number of electrons an atom would gain or lose when forming ionic bonds cannot be zero.</em>
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
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