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Troyanec [42]
3 years ago
9

A horse draws a sled horizontally across a snow-covered field. The coefficient of friction between the sled and the snow is 0.13

5, and the mass of the sled, including the load, is 195.9kg. If the horse moves the sled at a constant speed of 1.785m/s, what is the power needed to accomplish this?
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

P = 462.62 watts

Explanation:

The power needed to accomplish this can be calculated how

P = Fv

    Where F:  The force exerted by the horse

                v:  velocity

The force exerted by the horse is against friction force; how the movement is with constant velocity these forces must be equals, then

Fr = μN

    =μmg

    = (0.135)(195.9)(9.8)

    = 259.17 N

And the power is

P = (259.17)(1.785)

P = 462.62 watts

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Answer:

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Explanation:

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6 0
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A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
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Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

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V=41.96ft/s

b. t=(v-u) /g

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5 0
3 years ago
Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s
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Answer:

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