The neutral pH is 7. Less than 7 indicates an acid and more than 7 indicates a base (up to 14).
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NaCl - it's a salt (we can't measure the pH)
H2O - it can be an acid but also a base (the pH it is almost neutral,meaning close to 7 )
HF - it is a strong acid
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KOH
- it is a strong base (pH=14)
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↓
He needs to use HF (Hydrogen fluoride) to decrease the pH.
Answer:
Here, force=20N and displacement=10m
Work=Force×Displacement=20N×10m=200Nm
By definition, the momentum is given by:
p = m * v
Where,
m = mass
v = speed.
On the other hand,
F = m * a
Where,
m = mass
a = acceleration:
For the boy we have:
p1 = m * v
p1 = (F / a) * v
p1 = ((710) / (9.81)) * (0.50)
p1 = 36.19 Kg * (m / s)
For the girl we have:
p2 = m * v
p2 = (F / a) * v
p2 = ((480) / (9.81)) * (v)
p2 = 48.93 * v Kg * (m / s)
Then, we have:
p1 + p2 = 0
36.19 + 48.93 * v = 0
Clearing v:
v = - (36.19) / (48.93)
v = -0.74 m / s (negative because the velocity is in the opposite direction of the boy's)
Answer:
the girl's velocity in m / s after they push off is -0.74 m / s
Answer:
it goes up until we help it to but the moment we stop support it gets affected by gravity and goes back
Explanation:
Answer:
The velocity will be v = 22.1[m/s]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy is converted to kinetic energy. For this problem we will take the point with maximum potential energy when the body is 25 [m] high. By the time the height is zero, the potential energy will have been transformed into kinetic energy, and we can find the velocity of the body.
![Ep = m*g*h\\where:\\m = mass = 88.2[kg]\\h = elevation = 25[m]\\g = gravity = 9.81 [m/s^2]\\Ep = 88.2*25*9.81 = 21631.05[J]\\](https://tex.z-dn.net/?f=Ep%20%3D%20m%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%2088.2%5Bkg%5D%5C%5Ch%20%3D%20elevation%20%3D%2025%5Bm%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5CEp%20%3D%2088.2%2A25%2A9.81%20%3D%2021631.05%5BJ%5D%5C%5C)
Now we know that the energy will be transformed.
![Ek=Ep\\Ek=0.5*m*v^{2} \\where:\\v=velocity [m/s]\\v=\sqrt{\frac{Ek}{0.5*m} } \\v=\sqrt{\frac{21631.05}{0.5*88.2} } \\v=22.14[m/s]](https://tex.z-dn.net/?f=Ek%3DEp%5C%5CEk%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv%3Dvelocity%20%5Bm%2Fs%5D%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BEk%7D%7B0.5%2Am%7D%20%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B21631.05%7D%7B0.5%2A88.2%7D%20%7D%20%5C%5Cv%3D22.14%5Bm%2Fs%5D)
