Answer:
t_{out} = 
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
Density =mass/Volume
0.8=mass/60
Mass=0.8*60
Mass=48g
the equation of the tangent line must be passed on a point A (a,b) and
perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle,
its center is C(0,0). And we assume that the tangent line passes to the point
A(2.3).
</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.
<span>Let's find the equation of the line parallel to the radius.</span>
<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>
det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.
let's find the equation of the line perpendicular to this previous line.
let M a point which lies on the line. so MA.AC=0 (scalar product),
it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent
All of the above as it states that "<span>a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers"</span>
Gravity is the energy due to Earth pulling down on an object.
