Answer:
The final kinetic energy is 
Explanation:
From the question we are told that
The electric field is 
The charge on the object is 
The mass of the object is 
The distance moved by the object is 
The workdone on the object by the fields is mathematically represented as
![W = [qE + mg]d](https://tex.z-dn.net/?f=W%20%3D%20%20%5BqE%20%2B%20mg%5Dd)
Now this workdone is equivalent to the final kinetic energy so
![K = W = [qE + mg]d](https://tex.z-dn.net/?f=K%20%3D%20W%20%3D%20%20%5BqE%20%2B%20mg%5Dd)
substituting values
![K = W = [4.5*10^{-3 } *100 + 0.68 * 9.8]* 1](https://tex.z-dn.net/?f=K%20%3D%20W%20%3D%20%20%5B4.5%2A10%5E%7B-3%20%20%7D%20%2A100%20%20%2B%200.68%20%2A%209.8%5D%2A%201)
The weight of air resting on a surface, divided by the area
of the surface ... described in units of force per unit area ...
is called air pressure.
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.
Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
Balances and Scales
A balance compares an object with a known mass to the object in question. One example of a balance is the triple beam balance. The standard unit of measure for mass is based on the metric system and is typically denoted as kilograms or grams.
