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stealth61 [152]
3 years ago
14

A cylinder with rotational inertia I1 = 3.0 kg · m2 rotates clockwise about a vertical axis through its center with angular spee

d ω1 = 5.4 rad/s. A second cylinder with rotational inertia I2 = 1.3 kg · m2 rotates counterclockwise about the same axis with angular speed ω2 = 6.4 rad/s. If the cylinders couple so they have the same rotational axis, what is the angular speed of the combination (in rad/s)?
Physics
1 answer:
erastova [34]3 years ago
8 0

Answer:

ω = 1.83 rad/s clockwise

Explanation:

We are given:

I1 = 3.0kg.m2

ω1 = -5.4rad/s (clockwise being negative)

I2 = 1.3kg.m2

ω2 = 6.4rad/s  (counterclockwise being positive)

By conservation of the momentum:

I1 * ω1 + I2 * ω2 = (I1 + I2) * ω

Solving for ω:

\omega = \frac{I1 * \omega1 + I2*\omega2}{I1+I2}=-1.83rad/s

Since it is negative, the direction is clockwise.

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An 800 kg fishing boat going south at 12 m/s runs into a stopped pontoon boat (1400 kg). The boats stick together and move south
kykrilka [37]

Answer:

the velocity of the boats after the collision is 4.36 m/s.

Explanation:

Given;

mass of fish, m₁ = 800 kg

mass of boat, m₂ = 1400 kg

initial velocity of the fish, u₁ = 12 m/s

initial velocity of the boat, u₂ = 0

let the final velocity of the fish-boat after collision = v

Apply the principle of conservation of linear momentum for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

800 x 12    +   1400 x 0 = v(800 + 1400)

9600 = 2200v

v = 9600/2200

v = 4.36 m/s

Therefore, the velocity of the boats after the collision is 4.36 m/s.

7 0
3 years ago
Two particles move along an x axis. The position of particle 1 is given by x ! 6.00t 2 # 3.00t # 2.00 (in meters and seconds); t
s344n2d4d5 [400]

Answer:

15.6m/s

Explanation:

V1=\frac{dx}{dt}=\frac{d}{dt}(6t^{2}+3t+2) because the derivate of the position is the velocity

V1=12t+3

V2=20+\int\limits^_ {} \,-8t because the integral of the acceleration is the velocity

V2=20-4t^{2}

V1=V2 to see when the velocities of particles match

12t+3=20-4t^2

4t^2+12t-17=0 we resolve this with \frac{-b+-(\sqrt{b^{2} -4ac} )}{2a}

and we take the positif root

t=1.05 sec

if we evaluate the velocity (V1 or V2) the result is 15.6m/s

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3 years ago
The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the
vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r²             where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

I₂ = 2.1 x 10⁹ W/m²

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