Answer:x=23.4 cm
Explanation:
Given
mass of block ![m=0.5 kg](https://tex.z-dn.net/?f=m%3D0.5%20kg)
inclination ![\theta =30](https://tex.z-dn.net/?f=%5Ctheta%20%3D30)
coefficient of static friction ![\mu =0.35](https://tex.z-dn.net/?f=%5Cmu%20%3D0.35)
coefficient of kinetic friction ![\mu _k=0.25](https://tex.z-dn.net/?f=%5Cmu%20_k%3D0.25)
distance traveled ![d=77.3 cm](https://tex.z-dn.net/?f=d%3D77.3%20cm)
spring constant ![k=35 N/m](https://tex.z-dn.net/?f=k%3D35%20N%2Fm%20)
work done by gravity+work done by friction=Energy stored in Spring
![mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}](https://tex.z-dn.net/?f=mg%5Csin%20%5Ctheta%20d-%5Cmu%20_kmg%5Ccos%20%5Ctheta%20d%3D%5Cfrac%7Bkx%5E2%7D%7B2%7D)
![mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}](https://tex.z-dn.net/?f=mgd%5Cleft%20%28%20%5Csin%20%5Ctheta%20-%5Cmu%20_k%5Ccos%20%5Ctheta%20%5Cright%20%29%3D%5Cfrac%7Bkx%5E2%7D%7B2%7D)
![0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}](https://tex.z-dn.net/?f=0.5%5Ctimes%209.8%5Ctimes%200.773%5Cleft%20%28%20%5Csin%2030-0.25%5Ccos%2030%5Cright%20%29%3D%5Cfrac%7B35%5Ctimes%20x%5E2%7D%7B2%7D)
![x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B%5Cfrac%7B2%5Ctimes%200.5%5Ctimes%209.8%5Ctimes%200.773%28%5Csin%2030-0.25%5Ctimes%20%5Ccos%2030%29%7D%7B35%7D%7D)
![x=0.234 m](https://tex.z-dn.net/?f=x%3D0.234%20m)
![x=23.4 cm](https://tex.z-dn.net/?f=x%3D23.4%20cm)
Answer: vl = 2.75 m/s vt = 1.5 m/s
Explanation:
If we assume that no external forces act during the collision, total momentum must be conserved.
If both cars are identical and also the drivers have the same mass, we can write the following:
m (vi1 + vi2) = m (vf1 + vf2) (1)
The sum of the initial speeds must be equal to the sum of the final ones.
If we are told that kinetic energy must be conserved also, simplifying, we can write:
vi1² + vi2² = vf1² + vf2² (2)
The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:
vf1 = vi2 and vf2 = vi1
If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:
vf1 = 2.75 m/s vf2 = 1.5 m/s
I think it's A) Sunspots. I hope this helps:)
D-It will become a temporary magnet because the domains will easily realign.
Answer:
option (D) is correct.
Explanation:
According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.
the kinetic energy of a body is directly proportional to the square of the speed of the body.
As the kinetic energy change, the speed of the body also change.
Option (D) is correct.