Answer:
L = a 1,929 10⁴ m
a = 0.1 mm = 0.1 10⁻³ m, L = 1,929 m
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
Let's use trigonometry to find the breast
tan θ = y / x
As the angles are very small
tan θ = sin θ/ cos θ = sin θ = y / x
We replace
a y / L = m λ
L = a y / m λ
The red light has a wavelength of Lam = 700 nm = 700 10⁻⁹ m, in the third pattern it is m = 3
L = a 4.05 10⁻² / (3 700 10⁻⁹)
L = a 1,929 10⁴ m
To give a specific value we must know the width of the slit, suppose a value of a = 0.1 mm = 0.1 10⁻³ m
L = 1,929 m
M = 20.0kg, the mass of the block.
c = 1700 J/(kg-°C), the specific heat
ΔT = 25 - 15 = 10 °C = 10 K, the change in temperature.
The change in thermal energy is
Answer: 340 kJ (or 340,000 J)
1. clavicle = collarbone
2. vertebrae = backbone
3. scapula = shoulder blade
4. femur = thigh
5. humerus = upper arm
6. patella = kneecap
7. cranium = skull
8. tibia = lower leg
9. radius/ulna = forearm
10. phalanges = fingers/toes
Answer:
3.75 m/s
Explanation:
From the question given above, the following data were obtained:
Acceleration (a) = 1.5 m/s²
Initial velocity (u) = 0 m/s
Time (t) = 2.5 s
Final velocity (v) =?
a = (v – u) / t
1.5 = (v – 0) / 2.5
1.5 = v / 2.5
Cross multiply
v = 1.5 × 2.5
v = 3.75 m/s
Hence, the escape velocity of the squirrel is 3.75 m/s
Hello!!
Here we have a simple matter of conservation of energy. ME=PE+KE.
At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.
Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!
