Answer:
a) 11 m/s
b) 0.0564 s
Explanation:
Given:
m = 2100 kg
vi = 22 ..... m/s before collision
vf = 0 ......after collision to stop
Δs = 0.62 distance traveled after collision .. crumpling of truck
Part a

Part b

Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M
= 1.99 × 10³⁰ kg
Mass of the neutron star
M
= 2( M
)
M
= 2( 1.99 × 10³⁰ kg )
M
= ( 3.98 × 10³⁰ kg )
Radius of neutron star R
= 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω
.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM
= / R
² = mR
ω
²
ω
² = GM
= / R
³
ω
= √(GM
= / R
³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω
= √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω
= √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω
= √ 120831133.3636777
ω
= 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s
Answer:
191.36 N/m
Explanation:
From the question,
The Potential Energy of the safe = Energy of the spring when it was compressed.
mgh = 1/2ke²............... Equation 1
Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression
Making k the subject of the equation,
k =2mgh/e²................ Equation 2
Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m
Constant: g = 9.8 m/s²
Substitute into equation 2
k = 2(1100)(9.8)(0.0024)/0.52²
k = 51.744/0.2704
k = 191.36 N/m
Hence the spring constant of the heavy-duty spring = 191.36 N/m
Extensional stress. is your answer.