Answer:
i hope it will be useful for you
Explanation:
F=5.6×10^-10N
R=93cm=0.93m
let take m1 and m2 =m²
according to newton's law of universal gravitation
F=m1m2/r²
F=m²/r²
now we have to find masses
F×r²=m²
5.6×10^10N×0.93m=m²
5.208×10^-9=m²
taking square root on b.s
√5.208×10^-9=√m²
so the two masses are m1=7.2×10^-5
and m2=7.2×10^-5
I don't think the shaping of the beach or hill would be considered
weathering, especially since it says "as smaller particles are moved
away". This one is just talking about where the particles decide to
gather and make a shape.
'C' is really talking about weathering, where rocks are broken up into
stones, and the stones into smaller pieces that can be blown away
on the wind. THAT's weathering.
Yes, that is true. I hoped I helped!
Hello!
This is an example of an inelastic collision, where the two objects "stick" to each other after their collision. (The Goalkeeper CATCHES the puck).
We can write out the conservation of momentum formula:
m1vi + m2vi = m1vf + m2vf
Let:
m1 = mass of puck
m2 = mass of the goalkeeper
We know that the initial velocity of the goalkeeper is 0, so:
m1vi + m2(0) = m1vf + m2vf
m1vi = m1vf + m2vf
The final velocities will be the same, so:
m1vi = (m1 + m2)vf
Plug in the given values:
(0.16)(40)/ (0.16 + 120) = vf ≈ 0.0533 m/s
Using the equation for momentum:
p = mv
The object with the LARGER mass will have the greater momentum. Thus, the Goalkeeper has the largest momentum as p = mv; a greater mass correlates to a greater momentum since the velocity is the same between the two objects. The puck would have a momentum of p = (.16)(0.0533) = 0.008528 kgm/s, whereas the goalkeeper would have a momentum of
p = (120)(0.0533) = 6.396 kgm/s.
Yes,when they erupt witch leaves it unactive with makes it mountin
