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BigorU [14]
3 years ago
7

The correctly balanced equation for H 2O 2 → H 2O + O 2 is

Physics
1 answer:
antoniya [11.8K]3 years ago
3 0
2H2O + O2 -- 2H2O + O2 .
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Explain why engineers leave a gap in a concrete bridge (answer must relate to density)
seraphim [82]
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6 0
3 years ago
Plate Tectonic Theory
Vitek1552 [10]

Answer:

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Explanation:

6 0
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A person hits a 45-g golf ball. The ball comes down on a tree root and bounces
Verizon [17]

Answer:

Maximum height, h = 10 m          

Explanation:

It is given that,

Mass of golf ball, m = 45 g = 0.045 kg

The ball comes down on a tree root and bounces  straight up with an initial speed of 14.0 m/s.

We need to find the height the ball will rise  after the bounce. It is based on the conservation of energy such that,

\dfrac{1}{2}mv^2=mgh

h is maximum height raised by the ball

h=\dfrac{v^2}{2g}\\\\h=\dfrac{(14)^2}{2\times 9.8}\\\\h=10\ m

So, the ball will raised to a height of 10 meters.

5 0
3 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
Not a question but need a lot of tips
skad [1K]

DibujaZos da buenos tips e incluso el pinta con colores faber castel son económicos y el dibujo queda epico

8 0
3 years ago
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