Answer:
Explanation:
Hello,
Let's get the data for this question before proceeding to solve the problems.
Mass of flywheel = 40kg
Speed of flywheel = 590rpm
Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.
Time = 30s = 0.5 min
During the power off, the flywheel made 230 complete revolutions.
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + ω₂) / 2] × 0.5
But ∇θ = 230 revolutions
∇θ/t = (530 + ω₂) / 2
230 / 0.5 = (530 + ω₂) / 2
Solve for ω₂
460 = 295 + 0.5ω₂
ω₂ = 330rpm
a)
ω₂ = ω₁ + αt
but α = ?
α = (ω₂ - ω₁) / t
α = (330 - 590) / 0.5
α = -260 / 0.5
α = -520rev/min
b)
ω₂ = ω₁ + αt
0 = 590 +(-520)t
520t = 590
solve for t
t = 590 / 520
t = 1.13min
60 seconds = 1min
X seconds = 1.13min
x = (60 × 1.13) / 1
x = 68seconds
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + 0) / 2] × 1.13
∇θ = 333.35 rev/min
Answer:
Rebounce angle is 345°
Rebounce speed is 989.95m/s
Explanation:
Calculate the x component of the velocity of the bullet before impact by using the following relation:
Vbx= Vb Cos thetha
Here, is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°
Substituting
Vbx = Cos15 ×1400 = 1352.30m/s
Calculate the y component using the relation:
Vby = Vo Sin theta
Vby = sin 15° × 1400
Vby = 362.35m/s
The rebounce angle = 360 - incidence angle
Rebounce angle =( 360 - 15)° = 345°
The rebound speed V' = Vby - Vbx
V' = (1352.30 - 362.35)m/s
V' = 989.95 m/s
Answer: B,E,F
Explanation:
(B) the inner core is solid. (E) the outer core is liquid and moving. (F) Earth's core is composed of iron and nickel
