Answer:
There are five signs of a chemical change:
Color Change.
Production of an odor.
Change of Temperature.
Evolution of a gas (formation of bubbles)
Precipitate (formation of a solid)
Explanation:
I just went ahead and gave you the five signs of chemical change hoped it helped
30x30=900
The answer is 900 meters after 30 seconds
Answer:
Power of the string wave will be equal to 5.464 watt
Explanation:
We have given mass per unit length is 0.050 kg/m
Tension in the string T = 60 N
Amplitude of the wave A = 5 cm = 0.05 m
Frequency f = 8 Hz
So angular frequency 
Velocity of the string wave is equal to 
Power of wave propagation is equal to 
So power of the wave will be equal to 5.464 watt
Answer:
The first frequency of audible sound in the speed sound is
f = 662 Hz
Explanation:
vs = 344 m/s
x = 52 cm * 1 / 100m = 0.52m
The wave length is the distance between the peak and peak so
d = 2x
d = 2*0.52 m
d = 1.04 m
So the frequency in the speed velocity is
f = 1 / T
f = vs / x = 344 m/s / 0.52m
f ≅ 662 Hz
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)

q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.