Answer:E = hc/? = 4.41 x 10-19 J
Energy absorbed by each atom :
E (atom) = 2.205 x 10-19 J
Now Bond Energy of each molecule (B) = 3.98 x J
So, for each atom 1.99 x 10-19 J
So now
KE of each atom = E(atom) - B (atom)
= 2.15 x 10-19 J
B) gravitational to kinetic
Explanation:
The skydiver, when he is located at a certain height h above the ground, possesses gravitational potential energy, equal to:

where m is the mass of the skydiver, g is the gravitational acceleration and h is the height above the ground. As he falls, its height h decreases, while his speed v increases, so part of the gravitational potential energy is converted into kinetic energy, which is given by

so, we see that as v increases, the kinetic energy increases. Therefore the correct answer is
B) gravitational to kinetic
Answer:
The specific heat is 3.47222 J/kg°C.
Explanation:
Given that,
Temperature = 13°C
Temperature = 37°C
Mass = 60 Kg
Energy = 5000 J
We need to calculate the specific heat
Using formula of energy


Put the value into the formula


Hence, The specific heat is 3.47222 J/kg°C.
P=E/t
check physic Formula ..
answers is true
Answer:
t = 1.099 s
Explanation:
given,
constant speed = 2.51 m/s
height of balloon above ground = 3.16 m
time elapsed before it hit the ground = ?
Applying equation of motion to the compass



using quadratic formula to solve the equation


t = 1.099 s, -0.586 s
hence, the time elapses before the compass hit the ground is equal to 1.099 s.