Answer:
This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.
3]Explanation: This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.
4]Two moles of hydrogen peroxide H2O2 decomposes to produce two moles of water H2O and one mole of oxygen gas O2(g) , which then bubbles off
Answer:
137200000 watts or 137200 kilowatts
Explanation:
The formula for power is P= dhrg
Where P = Power in watts
d = density of water (1000 kg/m^3)
h = height in meters
r = flow rate in cubic meters per second,
g = acceleration due to gravity of 9.8 m/s^2,
Plugging in the known values,
we get
P = 1000 kg/m^3 * 80 m * 175 m^3/s * 9.8 m/s^2
P = 80000 kg/m^2 * 175 m^3/s * 9.8 m/s^2
P = 14000000 kg m/s * 9.8 m/s^2
P = 137200000 kg m^2/s^3
P = 137200000 watts or 137200 kilowatts
The above figure assumes 100% efficiency which is impossible. A good efficiency would be 90% so the actual power available would be close to 0.90 * 137200 = 123480 kilowatts
Answer:
38.4 m/s
Explanation:
a) at t = 3.2s. ![x = 6 * 3.2^2 = 61.44 m](https://tex.z-dn.net/?f=x%20%3D%206%20%2A%203.2%5E2%20%3D%2061.44%20m)
b) at t = 3.2 + Δt. ![x = 6*(3.2 + \Delta t)^2](https://tex.z-dn.net/?f=x%20%3D%206%2A%283.2%20%2B%20%5CDelta%20t%29%5E2)
c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt
![v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B%5CDelta%20x%7D%7B%5CDelta%20t%7D%20%3D%20%5Cfrac%7Bx_2%20-%20x_1%7D%7B%5CDelta%20t%7D)
![v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B6%2A%283.2%20%2B%20%5CDelta%20t%29%5E2%20-%2061.44%7D%7B%5CDelta%20t%7D)
![v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B6%283.2%5E2%20%2B%206.4%5CDelta%20t%20%2B%20%5CDelta%20t%5E2%29%20-%2061.44%7D%7B%5CDelta%20t%7D)
![v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B61.44%20%2B%2038.4%5CDelta%20t%20%2B%20%5CDelta%20t%5E2%20-%2061.44%7D%7B%5CDelta%20t%7D)
![v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B%5CDelta%20t%2838.4%20%2B%20%5CDelta%20t%29%7D%7B%5CDelta%20t%7D)
![v = 38.4 + \Delta t](https://tex.z-dn.net/?f=%20v%20%3D%2038.4%20%2B%20%5CDelta%20t)
As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s
Its very dense. Hey, are you homeschooled?
Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151