Answer:
nope its a myth don't worry :)
These are two questions and two answers
Question 1.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 9.11 × 10⁻³¹ kg
b) λ = 3.31 × 10⁻¹⁰ m
c) c = 3.00 10⁸ m/s
d) s = ?
<u>2) Formula:</u>
The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Solve for s:
Substitute:
- s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s
To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s
- s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³
Answer: s = 7.33 × 10 ⁻³ c
Question 2.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 45.9 g (0.0459 kg)
b) s = 70.0 m/s
b) λ = ?
<u>2) Formula:</u>
Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Substitute:
- λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m
As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.
Answer: 2.06 × 10 ⁻³⁴ m.
A Liquid called Surface Tension.
Answer:
48.67 seconds
Explanation:
From;
1/[A] = kt + 1/[A]o
[A] = concentration at time t
t= time taken
k= rate constant
[A]o = initial concentration
Since [A] =[A]o - 0.75[A]o
[A] = 0.056 M - 0.042 M
[A] = 0.014 M
1/0.014 = (1.1t) + 1/0.056
71.4 - 17.86 = 1.1t
53.54 = 1.1t
t= 53.54/1.1
t= 48.67 seconds
Hence,it takes 48.67 seconds to decompose.
This is a tricky question. All that matters are ratios of percentages, not percentages themselves. So no one should directly compare 27.2 with 42.9. We must and shall compare the ratios (27.2 to 72.8) and (42.9 to 57.1).
Take them both down to 1 to and see what happens.
Working out the formulas knowing atomic masses is a bit beside the point; this is how people first DISCOVERED the idea of atomic mass.
A
Carbon Oxygen
27.2g 72.8g (100-27.2)
Moles 27.2/12 72.8/16
2.27 4.55
Ratio 1 2
Do the same with the other
