Answer:
Final velocity = 7.677 m/s
KE before crash = 202300 J
KE after crash = 182,702.62 J
Explanation:
We are given;
m1 = 1400 kg
m2 = 4700 kg
u1 = 17 m/s
u2 = 0 m/s
Using formula for inelastic collision, we have;
m1•u1 + m2•u2 = (m1 + m2)v
Where v is final velocity after collision.
Plugging in the relevant values;
(1400 × 17) + (4700 × 0) = (1400 + 1700)v
23800 = 3100v
v = 23800/3100
v = 7.677 m/s
Kinetic energy before crash = ½ × 1400 × 17² = 202300 J
Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
Answer:
option (E) 1,000,000 J
Explanation:
Given:
Mass of the suspension cable, m = 1,000 kg
Distance, h = 100 m
Now,
from the work energy theorem
Work done by the gravity = Work done by brake
or
mgh = Work done by brake
where, g is the acceleration due to the gravity = 10 m/s²
or
Work done by brake = 1000 × 10 × 100
or
Work done by brake = 1,000,000 J
this work done is the release of heat in the brakes
Hence, the correct answer is option (E) 1,000,000 J
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
Answer: 10m/s
Explanation: Since a single vector with length of 1cm expresses 5m/s, the vector which has a length doubled from the original vector should have the speed which is also doubled.
