Answer:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² = 
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo = ![v_{oy}² /2 g y- 0 = 10.0²/2 9.8 y - 0 = 5.10 m The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system y₂ = 5.1 + 44 y₂ = 49.1 m Let's use the other equation to find the time [tex]v_{y}](https://tex.z-dn.net/?f=v_%7Boy%7D%C2%B2%20%2F2%20g%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y-%200%20%3D%2010.0%C2%B2%2F2%209.8%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y%20-%200%20%3D%205.10%20m%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%0A%3C%2Fp%3E%3Cp%3EThe%20height%20from%20the%20ground%20is%20the%20height%20that%20rises%20from%20the%20reference%20system%20plus%20the%20depth%20of%20the%20ground%20from%20the%20reference%20system%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%205.1%20%2B%2044%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%2049.1%20m%0A%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ELet%27s%20use%20the%20other%20equation%20to%20find%20the%20time%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20%20%5Btex%5Dv_%7By%7D)
= - g t
t = / g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 = ² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s
Answer:
1.170*10^-3 m
3.23*10^-32 m
Explanation:
To solve this, we apply Heisenberg's uncertainty principle.
the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"
Δp * Δx = h/4π
m(e).Δv * Δx = h/4π
If we make Δx the subject of formula, by rearranging, we have
Δx = h / 4π * m(e).Δv
on substituting the values, we have
for the electron
Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 5.67*10^-31
Δx = 1.170*10^-3 m
for the bullet
Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 0.021
Δx = 3.23*10^-32 m
therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet
The reactants are <em>Calcium</em>, <em>Hydrogen</em> and <em>Oxygen</em>.
We need 1 atom of <em>Calcium</em>, 4 atoms of <em>Hydrogen</em> and 2 atoms of <em>Oxygen</em>.
Hope this helps.
r3t40
Answer:
D. transparent.
Explanation:
A material that transmits nearly all the light in a ray because it offers little resistance to the light is <u>transparent.</u>
A transparent material allows light to pass through them with little or no resistance enabling them see-through. A material that transmits nearly all the light rays that pass through it because it offers little resistance to the light is TRANSPARENT. Examples of transparent materials are water, glass (flint and crown), air, and diamond.
Gas pressure will also increase then.., if volume is kept constant
