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Cerrena [4.2K]
2 years ago
7

Three joggers are running along straight lines as follows: Jogger A, with a mass of 55.2kg, is traveling along the line y=6.00m

at 3.45 m/s in the positive x-direction. Jogger B, with a mass of 62.4kg, is traveling at 4.23 m/s along the line x=3.00m in the positive y-direction. Jogger C, with a mass of 72.1kg, is traveling at 4.75 m/s along the line x=-5.00m in the negative y-direction. What is the total counterclockwise angular momentum of the three joggers about the origin?
Physics
1 answer:
frutty [35]2 years ago
7 0

Answer:

L = - 1361.591 k Kgm/s

Explanation:

Given

mA = 55.2 Kg

vA = 3.45 m/s

rA = 6.00 m

mB = 62.4 Kg

vB = 4.23 m/s

rB = 3.00 m

mC = 72.1 Kg

vC = 4.75 m/s

rC = - 5.00 m

then we apply the equation

L =  (mv x r)

⇒  LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s

⇒  LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s

⇒  LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s

Finally, the total counterclockwise angular momentum of the three joggers about the origin is

L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k  Kgm/s

L = - 1361.591 k Kgm/s

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maxonik [38]

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We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

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2 years ago
If a 110 kg go-cart traveling at a velocity of 13.41 m/s has a collision with an impulse of 615 Nxs, what is the
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Answer:

5.59 m/s

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We are given;

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\Acceleration(a) = 615/110 = 5.591 m/s²

Now, using Newton's first law of motion, we can find acceleration (a). Thus;

v = u + at

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2 years ago
A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
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Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

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Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

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Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

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Ф = 93750 rad

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1 rad = 0.159155 rev,

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<em />

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

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Artyom0805 [142]

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