Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:

Our values are given by,

As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:


Skate 2:


Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:




Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Any photos so i can help you with that?
Answer:
7 / 1
Explanation:
The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength
Ax / Ay = λx / λy = 1 / 7
λy / λx = 7 / 1
120/60= 2m per s
220/55= 4m per s
720/80= 9m per s
Fastest run: 3rd run
Speed: 9m per s
Slowest: 1st run
Speed: 2m per s
Gravitational force = G · m₁ · m₂ / r²
' G ' is the 'universal gravitational constant, 6.673 x 10⁻¹¹ m³/kg-s²
m₁ = 60 kg
m₂ = 120 kg
r = 5 m
F = (6.673 x 10⁻¹¹ m³/kg-sec²) · (60 kg) · (120 kg) / (5m)²
F = (6.673 x 10⁻¹¹ · 60 · 120 / 25) · (m³ · kg · kg / kg · s² · m²)
F = 1.922 x 10⁻⁸ (kg · m / s²)
<em>F = 1.922 x 10⁻⁸ Newton</em> (about 0.0000000692 ounce of force)