Answer:
a) 0.049 m
b) Yes, increase
Explanation:
Draw a free body diagram.
In the y direction, there are three forces acting on the feeder. Two vertical components of the tension forces in each rope pulling up, and weight force pulling down.
Apply Newton's second law to the feeder in the y direction.
∑F = ma
2Ty − mg = 0
Ty = mg/2
Let's say the distance the rope sags is d. The trees are 4m apart, so the feeder is 2m horizontally from either tree. Using Pythagorean theorem, we can find the length of the rope on either side:
L² = 2² + d²
L = √(4 + d²)
Using similar triangles, we can write a proportion using the forces and distances.
Ty / T = d / L
Substitute:
(mg/2) / T = d / √(4 + d²)
Solve for d:
Td = mg/2 √(4 + d²)
T² d² = (mg/2)² (4 + d²)
T² d² = (mg)² + (mg/2)² d²
(T² − (mg/2)²) d² = (mg)²
d² = (mg)² / (T² − (mg/2)²)
d = mg / √(T² − (mg/2)²)
Given m = 2.4 kg and T = 480 N:
d = (2.4) (9.8) / √(480² − (2.4×9.8/2)²)
d = 0.049 m
b) If a bird lands on a feeder, this will increase the tension in the rope to support the bird's weight.
Answer:
C. Consider projectile motion. If an object starts out at some angle and only vertical forces (gravity) are present, the horizontal speed remains constant thruout the motion.
The work done in lifting the hamburger is equal to the increase in gravitational potential energy of the hamburger, given by
where
m=0.1 kg is the mass of the hamburger
is the gravitational acceleration
is the increase in height of the hamburger
Substituting numbers into the equation, we find
So, the correct answer is
(3) 0.3 J
I got transitional and rotational motion on my test
red having the frequency of 4.62 and at its limit 4.29 has the lowest frequency. It is the lowest before orange which has 5.00.