Answer:
a) 11 m/s
b) 0.0564 s
Explanation:
Given:
m = 2100 kg
vi = 22 ..... m/s before collision
vf = 0 ......after collision to stop
Δs = 0.62 distance traveled after collision .. crumpling of truck
Part a
Part b
Answer:
#_photon = 5 10²⁰ photons / s
Explanation:
For this exercise let's calculate the energy of a single quantum of energy, use Planck's law
E = h f
c= λ f
E = h c / λ
λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m
Let's calculate
E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹
E₀ = 19.89 10⁻²⁰ J
This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w
#_photon = P / E₀
#_photon = 100 / 19.89 10⁻²⁰
#_photon = 5 10²⁰ photons / s
Answer:
-48 N
Explanation:
mass of door (m) = 4 kg
acceleration of the door = 12 m/s^{2}
force exerted by the person = 48 N
From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N
39.2 J
Explanation:
Step 1:
To find the potential energy the following formula is used.
Potential Energy = m × g × h
Where,
m = Mass
g = Acceleration due to gravity
h = Height
Step 2:
Here m = 4 kg, g = 9.8 m/s², h = 1 m
Potential Energy = ( 4 × 9.8 × 1)
= 39.2 J
