Explanation:
Given that,
Wavelength = 6.0 nm
de Broglie wavelength = 6.0 nm
(a). We need to calculate the energy of photon
Using formula of energy



(b). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy


Put the value into the formula


(c). We need to calculate the energy of photon
Using formula of energy



(d). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy


Put the value into the formula


Hence, This is the required solution.
Answer:
c. Joints allow the roadway to expand and contract as cars put force on the bridge
Explanation:
The reasons why joints are allowed on roadway is to accommodate the contraction and expansion of the road as cars put force on them.
- Most materials used in making roadways are susceptible to expansion and contraction.
- When a measure of force is applied their length either increases or decreases depending on the type of force.
- To accommodate these changes, joints are placed in roadways
Answer:

Explanation:
Since the hoop is rolling on the floor so its total kinetic energy is given as

now for pure rolling condition we will have

also we have

now we will have


now by work energy theorem we can say



now solve for final speed

That was sun as some smaller masses formed planets and other remaining formed sun
I hope it helps! good luck in chem!