Answer: 585 J
Explanation:
We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

Answer:
a) y = 2.4 x 10⁻³ m = 0.24 cm
b) y = 3.2 x 10⁻³ m = 0.32 cm
Explanation:
The formula of Young's Double Slit experiment will be used here:

where,
y = distance between dark spots = ?
λ = wavelength
L = distance of screen = 2 m
d = slit width = 4 x 10⁻⁴ m
a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:

<u>y = 2.4 x 10⁻³ m = 0.24 cm</u>
<u></u>
a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:

<u>y = 3.2 x 10⁻³ m = 0.32 cm</u>
Given :
Two forces act on a 6.00-kg object. One of the forces is 10.0 N.
Acceleration of object 2 m/s².
To Find :
The greatest possible magnitude of the other force.\
Solution :
Let, other force is f.
So, net force, F = 10 + f.
Now, acceleration is given by :

Therefore, the greatest possible magnitude of the other force is 2 N.
Hence, this is the required solution.
It's b, because the more force an object it is given the harder it will be for it to slow down.
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km