Answer:
The ball will be at 700 m above the ground.
Explanation:
We can use the following kinematic equation
.
where y(t) represent the height from the ground. For our problem, the initial height will be:
.
The initial velocity:
,
take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:
So, the equation for our problem its:
.
Taking t=6 s:
.
.
.
.
.
So this its the height of the ball 6 seconds after being thrown.
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
