To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.
PART A) Normal Force.
Here,
Normal reaction of the ring is N and velocity of the ring is v
PART B) Acceleration
Negative symbol indicates deceleration.
<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>
Answer:
F = 1263.03 N
Explanation:s
given,
mass of the disk thrower = 100 Kg
mass of the disk = 2 Kg
angular speed of the disk = 4 rev/s
arm outstretched = 1 m
centripetal force of the disk in the circular path
F = m ω² r
ω = 4 x 2 x π
ω = 25.13 rad/s
F = m ω² r
F = 2 x 25.13² x 1
F = 1263.03 N
hence, centripetal force equal to the F = 1263.03 N
Answer:
Tuesday bc instead of running he/she was walking bc he/she might not have as much energy
Explanation:
Answer:
D. Calculate the area under the graph.
Explanation:
The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)
You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.
Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.
Theories have both an explanatory an a predictive function. True
