Answer:
<h2>154.73N</h2>
Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
There are at least two forces on it, and there could be more.
Vertical forces:
-- gravity, directed downward
-- buoyant force, directed upward
These two forces must be exactly equal, so that the net
vertical force on the raft is zero. Otherwise, it would be
accelerating either up or down.
Horizontal forces:
We know that the net horizontal force on the raft is zero.
Otherwise, it would be accelerating horizontally.
But we don't know if there are actually no horizontal forces
at all, or a balanced group of horizontal forces, that add up
to a net force of zero.
Answer:
g = 0.85 m
Explanation:
g = 
were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x 
N
), M is the mass of the earth (5.972 x 
kg), and h is the distance of meteoroid to the earth.
h = 3.40 x R
= 3.40 x 6371 km
h = 21661.4 km
= 21661400 m
Thus,
g = 
= 
= 0.84944
g = 0.85 m
The acceleration due to the Earth's gravitation is 0.85 m.
