Answer:
0.7515875 eV

Explanation:
f = Maximum frequency = 
h = Planck's constant = 
W = Work function = 2.52 eV
Converting to Joules

Maximum photon energy is given by

Maximum Kinetic energy is given by

Converting to eV

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

The range of frequencies for which no electrons are ejected is

Answer:
Yes
Explanation:
The given parameters are;
The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)
The angle in which the fastball is hit, θ = 22°
The distance of the field = 96 m (315 ft)
The range of the projectile motion of the fastball is given by the following formula

Where;
g = The acceleration due to gravity = 9.81 m/s², we have;

Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.
Answer:
1371.4watt
Explanation:
from power=energy/time
BUT energy=force times distance
Answer:
The value of spring constant is 266.01 
Explanation:
Given:
Mass of pellet
kg
Height difference of pellet rise
m
Spring compression
m
From energy conservation law,
Spring potential energy is stored into potential energy,

Where
spring constant, 



Therefore, the value of spring constant is 266.01 
<span>you can look at magnesium, it can react with oxygen to form oxides. (chemical) it is malleable and a solid at room temperature. (physical)
</span><span>to measure its density, the mass and volume can be worked out and from this density too. look up the equation, it is quite easy :)
</span><span>physical changes -- it can be melted, and oxidized
</span><span>the chemical changes of oxidation magnesium looses electrons to form oxides, this is a chemical reaction- chemical change..--- use to get the density use (rho) or density D = M/V</span>