I believe this is electron degeneracy, because the star is essentially having too many reactions too fast and collapses in on itself eventually.
Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N
Answer:
![L_{o}=0.1224m](https://tex.z-dn.net/?f=L_%7Bo%7D%3D0.1224m)
Explanation:
Given data
Force F=2 N
Length L=17 cm = 0.17 m
Spring Constant k=42 N/m
To find
Relaxed length of the spring
Solution
From Hooke's Law we know that
![F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m](https://tex.z-dn.net/?f=F_%7Bspring%7D%3Dk_%7Bs%7D%7Cs%7C%5C%5CF_%7Bspring%7D%3Dk_%7Bs%7D%28L-L_%7Bo%7D%29%5C%5C%202N%3D%2842N%2Fm%29%280.17m-L_%7Bo%7D%29%5C%5C2%3D7.14-42L_%7Bo%7D%5C%5C-42L_%7Bo%7D%3D2-7.14%5C%5C42L_%7Bo%7D%3D5.14%5C%5CL_%7Bo%7D%3D%285.14%2F42%29%5C%5CL_%7Bo%7D%3D0.1224m)
Answer:
Initial velocity, U = 4.5m/s
Explanation:
Given the following data;
Final velocity, v = 12m/s
Time, t = 5 seconds
Acceleration, a = 1.5m/s²
To find the initial velocity, we would use the first equation of motion.
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
12 = U + 1.5*5
12 = U + 7.5
U = 12 - 7.5
Initial velocity, U = 4.5m/s