Answer:
The jug drowns because the density of the jug is more than that of the density of water.
1. Roll a ball across a table into an object
2. Drop something
Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
The answer is D. 32 m.
The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
![v= \frac{d}{t}](https://tex.z-dn.net/?f=v%3D%20%5Cfrac%7Bd%7D%7Bt%7D%20)
⇒
![d=v*t](https://tex.z-dn.net/?f=d%3Dv%2At)
It is given:
![v = \frac{1.6m}{0.5s} = \frac{1.6m*2}{0.5s*2}= \frac{3.2m}{1s} = 3.2 m/s](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7B1.6m%7D%7B0.5s%7D%20%3D%20%5Cfrac%7B1.6m%2A2%7D%7B0.5s%2A2%7D%3D%20%5Cfrac%7B3.2m%7D%7B1s%7D%20%20%3D%203.2%20m%2Fs)
t = 10 s
d = ?
So:
Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV
Explanation: To solve this problem we have to use the relationship given by De Broglie as:
λ =p/h where p is the momentum and h the Planck constant
if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m
Finally we have:
eΔV=p^2/2m= h^2/(2*m*λ^2)
replacing we obtained the above values.