Answer:
633nm
Explanation:
Given the following :
Number of lines per centimeter(N) = 7000
Angle θ = 62.4°
Order (n) = 2
If grating element = d
Wavelength (λ) = (d* SinΘ) / 2
If number of lines = 7000 per cm
Converting to metre :
100 cm = 1m
7000 lines per 1 cm
Number of lines per m:
7000 lines * 100 = 700,000 lines per meter
Recall :
d = reciprocal of N
d = 1 / 700,000
d = 0.00000142857
Substituting into (λ) = (d* SinΘ) / 2
λ = (0.00000142857 * Sin 62.4°) / 2
λ = 0.00000126600 / 2
λ = 0.000000633002
λ = 0.000000633
λ = 633 × 10^-9 m = 633nm
<h3>
Answer:</h3>
B. (PE)beginning = (KE)end
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of an object is 72.0 kg
- Velocity of the body before it hits the ground is 79.0 m/s
We are required to determine the relationship between the potential energy and kinetic energy before the fall.
- When an object is at the highest point, it has maximum potential energy and minimum kinetic energy.
- This is because potential energy is directly proportional to the height of an object above the earth's surface.
- On the other hand, when an object attains the highest speed it has maximum kinetic energy and minimum kinetic energy.
In this case;
- The velocity of the object when hitting the ground is maximum and thus the object will have maxim,um kinetic energy.
- As the object falls towards the ground the potential energy is being converted to kinetic energy.
- Therefore, the potential energy at the beginning will be equal to the kinetic energy at the end when the object is on the ground.
- We can therefore, conclude that, (PE)beginning = (KE)end
Force = mass x acceleration
so to find acceleration you would do force divided by mass
Answer
given,
t = 3 s
we know,
position of the particle
integrating both side
Position of the particle at t= 3 s
x = 182.98 ft
Distance traveled by the particle in 3 s is equal to 182.98 ft
now, particle’s acceleration
at t= 3 s
a = 2.98 ft/s²
acceleration of the particle is equal to 2.98 ft/s²
Answers:
A) 
B) 22. 5 N
Explanation:
A) Since this situation is related to uniform circular motion, the centripetal acceleration 
of the child is calculated by:
Where:
is the speed
is the child's distane to the center (the radius)
Hence:
B) Knowing th centripetal acceleration and the mass 
of the child, we can calculate the net horizontal force 
by:
