Answer:
Explanation:
The radiated power can be given in terms of the wavelength as follows:
where,
Radiated Power = 1.2 x 10⁻¹⁷ W
n = no. of photons = ?
h = plank's constant = 6.625 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 500 nm = 5 x 10⁻⁷ m
t = time
Therefore,
1=inelastic and 2=elastic hope this helps
Answer:
As the trailer leaks sand at a constant rate through a hole in its bottom, its acceleration increases at a steady rate. (Option A)
Explanation:
F = ma
where;
F is the force applied on the truck
m is the mass of the truck
a is the acceleration of the truck
If the driving force on the truck remain constant, then;
m₁a₁ = m₂a₂
a₁ is the initial acceleration
a₂ is the final acceleration
m₁ is the initial mass of the truck and loaded sand,
m₂ is the final mass of the truck and loaded sand, which decreases as the sand leaks at a constant rate.
a₂ = (m₁a₁ )/(m₂)
since m₁ > m₂, a₂ > a₁
Therefore, as the trailer leaks sand at a constant rate through a hole in its bottom, its acceleration increases at a steady rate.
Explanation:
When we draw the diagram of the motion we get a right angled triangle.
We have an angle of 30° and a base of 17.3M.
We will use trigonometric ratios to solve this.
We will use tan.
Tan = Opposite / adjacent
Tan 30° = h / 17.3
h = Tan 30 × 17.3
h = 0.5774 × 17.3 = 9.99 M
or 9.99 can be roundoffed with 10m
Answer:
16.67 rad/s
Explanation:
if you read the question, you would notice that some requirements are missing, while searching online, the complete question can be found here:
https://www.chegg.com/homework-help/questions-and-answers/frictionless-pulley-modeled-080-solid-cylinder-030-m-radius-rope-going--tension-rope-10-n--q3499646
and here is the complete question:
"a frictionless pulley which can be modeled as a 0.80 solid cylinder with a 0.30 m radius has a rope going over it. The tension in the rope is 10 N on the right side and 12 N on the left side. What is the angular acceleration of the pulley?
"
mass (m) = 0.8 kg
radius (r) = 0.3 m
tension on left side (T1) = 10 N
tension on right side (T2) = 12 N
angular acceleration = torque / moment of inertia
where
torque = (12-10) x 0.3 = 0.6 Nm
- moment of inertia = 0.5m
moment of inertia = 0.5 x 0.8 x = 0.036 kg
therefore
angular acceleration = 0.6/0.036 = 16.67 rad/s