1 atm corresponds to 760 mmHg, so we can set up a simple proportion to find how many atmospheres correspond to 570 mmHg:

and from this, we find
Answer:
a) the one with a lower orbit b) the one with a higher orbit
Explanation:
Let's consider orbital mechanics. To get an object in orbit, we need it to fall to earth parallel to the earth's surface. To understand it easily imagine a projectile thrown horizontally further and further away, at one point, the projectile hits the cannon from behind. Considering there is no wind resistance, that would be a projecile in orbit.
In other words, the circular orbits of some objects around a massive body are due to the equality between centrifugal acceleration and gravity acceleration.
.
so the velocity is

where "G" is the gravitational constant, "M" the mass of the massive body and "r" the distance between the object and the center of gravity of mass M. As you can note, if "r" increase, "v" decrease.
The orbital period of any object in orbit is

where "a" is length of semi-major axis (a = r in circular orbits). So if "r" increase, "T" increase.
Work is calculated as the product of Force, Distance, and
angular motion. In this case, the work done by gravity is perpendicular to the motion
of the cart, so θ = 90°
and W=Fdcosθ
W=35.0 N x 20.0 m x cos90
W=0 J
This means that work done perpendicular to the direction of
the motion is always zero.
Answer:
F > W * sin(α)
Explanation:
The force needed for the box to start sliding up depends on the incline (α).
The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.
These forces can be decomposed on their normal and tangential (to the slide plane) components.
The weight will be split into
Wn = W * cos(α) (in normal direction)
Wt = W * sin(α) (in tangential direction)
The normal reaction will be alligned with the normal axis, and will be equal to -Wn
N = -W* cos(α) (in normal direction)
To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger
F > |-W * sin(α)| (in tangential direction)