The mass m of the object = 5.25 kg
<h3>Further explanation</h3>
Given
k = spring constant = 3.5 N/cm
Δx= 30 cm - 15 cm = 15 cm
Required
the mass m
Solution
F=m.g
Hooke's Law
F = k.Δx
Answer:
Friction can be minimized by using lubricants like oil and grease and by using ball bearing between machine parts. A substance that is introduced between two surfaces in contact, to reduce friction, is called a lubricant. Fluid friction can be minimized by giving suitable shapes to the objects moving in the fluids.
Explanation:
hope it helps
Are there any options??
I would have to say metal of course but without options I can't assume anything
Assuming this is an elastic collision, they go in the direction of the object with more mass
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.
