Answer:
Explanation:
Given that
Mass of bowling ball M1=7.2kg
The radius of bowling ball r1=0.11m
Mass of billiard ball M2=0.38kg
The radius of the Billiard ball r2=0.028m
Gravitational constant
G=6.67×10^-11Nm²/kg²
The magnitude of their distance apart is given as
r=r1+r2
r=0.028+0.11
r=0.138m
Then, gravitational force is given as
F=GM1M2/r²
F=6.67×10^-11×7.2×0.38/0.138²
F=9.58×10^-9N
The force of attraction between the two balls is
F=9.58×10^-9N
True because well it’s moving fast lol sometimes ur eyes have a hard time following its speed
My calculator is about 1cm thick, 7cm wide, and 13cm long.
Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .
The question wants me to assume that the density of my calculator
is about the same as the density of water. That doesn't seem right
to me. I could check it easily. All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically.
I won't do that. I'll accept the assumption.
If its density is actually 1 g/cm³, then its mass is about 91 grams.
The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.
My result of 91 grams is about 100 grams ... about 10² grams.
Your results could be different.
No, no me habla espanol. yo soy ingles
The runner has initial velocity vector
and acceleration vector
so that her velocity at time 
is
She runs directly east when the vertical component of 
is 0:
It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time would be
so that after 10.4 s, her position would be
which is 19.9 m away from her starting position.
