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mrs_skeptik [129]
3 years ago
12

A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine

and hit a stationary green bumper car of mass 150 kg. The red, blue, and green bumper cars all combine.
A. 0.93 m/s
B 1.2 m/s
C 1.7 m/s
D 2.7 m/s
Physics
2 answers:
GuDViN [60]3 years ago
8 0

Answer:

B. 1.2m/s

Explanation:

as per the law of conservation of momentum,

initial momentum= final momentum

considering collision 1 : Mr * Vr + ( Mb * Vb) = ( Mr + Mb ) Vrb

where Vrb is the final velocity after 1st collision with which red and blue car moves.

therefore,  225* 3 + 0 = ( 225 + 180 ) Vrb

Vrb = 675/405

       = 1.67m/s

now collision 2 : ( Mr+ Mb) Vrb + MgVg= ( Mr+ Mg+ Mb ) V'

where V' is the total speed

675 + 0 = 555V'

V' = 675/555

   = 1.2 m/s

klemol [59]3 years ago
6 0
I think the answer for the question above its            b 1.2
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You accidentally slide a glass of milk off a table that is 0.9 m tall. How long does it take the milk to hit the ground.
labwork [276]

Fine, lets do a retry of this.

Δd = -0.9m

v₁ = 0

v₂ = ?

a = -9.8 m/s²

Δt = ?

We can use the following kinematic equation and solve for Δt.

                             Δd = v₁Δt + 0.5(a)(Δt)²

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3 years ago
Calculate the moment of inertia for each scenario: (a) An 80.0 kg skater is approximated as a cylinder with a 0.140 m radius. (b
Zina [86]

Answer:

a) the moment of inertia is 0.784 Kg*m²

b) the moment of inertia is with arms extended is 1.187 Kg*m²

c) the angular velocity in scenario (b) is 4.45 rad/s

Explanation:

The moment of inertia is calculated as

I = ∫ r² dm

since

I = Ix + Iy

and since the cylinder rotates around the y-axis then Iy=0 and

I = Ix = ∫ x² dm

if we assume the cylinder has constant density then

m = ρ * V = ρ * π R²*L = ρ * π x²*L

therefore

dm = 2ρπL* x dx

and

I = ∫ x² dm = ∫ x² 2ρπL* x dx = 2ρπL∫ x³ dx = 2ρπL (R⁴/4 - 0⁴/4) = ρπL R⁴ /2 =  mR² /2

therefore

I skater = mR² /2 = 80 Kg * (0.140m)²/2 = 0.784 Kg*m²

b) since the arms can be seen as a thin rod

m = ρ * V = ρ * π R²*L = ρ * π R²*x

dm =ρ * π R² dx

I1 = ∫ x² dm = ∫ x² * ρ * π R² dx = ρ * π R²*∫ x² dx = ρ * π R²* ((L/2)³/3 - (-L/2)³/3)

= ρ * π R²*2*L³/24 = mL²/12

therefore

I skater 2 = I1 + I skater =  mL²/12 + mR² /2= 8 Kg* (0.85m)²/12 +(80-8) Kg * (0.140m)²/2 = 1.187 Kg*m²

c)  from angular momentum conservation

I s2 * ω s2 = I s1 * ω s1

thus

ω s2 = (I s1 / I s2 )* ω s1 /= (0.784 Kg*m²/1.187 Kg*m²) * 6.75 rad/s = 4.45 rad/s

4 0
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