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kirill [66]
3 years ago
13

A cart for hauling ore out of a gold mine has a mass of 435 kg, including its load. the cart runs along a straight stretch of tr

ack that is sloped 4.77° from the horizontal. a donkey, trudging along and to the side of the track, has the unenviable job of pulling the cart up the slope with a 438-n force for a distance of 193 m by means of a rope that is parallel to the ground and makes an angle of 13.3° with the track. the coefficient of friction for the cart\'s wheels on the track is 0.0167. use g = 9.81 m/s2. find the work that the donkey performs on the cart during this process.
Physics
1 answer:
marta [7]3 years ago
4 0
What are the answer choices 

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An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e
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Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

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A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0
3 years ago
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