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3 years ago

How much work output can a 150hp snowmobile do in 15 seconds if it is 18 % efficiency?

Physics

1 answer:

Sav [38]3 years ago

3 0

The work output is $3.02\cdot 10^5 J$

Explanation:

The power of the snowmobile is

$P=150 hp$

Keeping in mind that

$1 hp = 746 W$

We can convert it into Watts:

$P=(150)(746)=111,900 W$

The energy used by the snowmobile can be found as follows:

$E=Pt$

where

P = 111,900 W is the power used

t = 15 s is the time interval

Substituting,

$E=(111,900)(15)=1.68\cdot 10^6 J$

However, the snowmobile is 18% efficient: this means that only 18% of this energy is converted into useful work. Therefore, the work output is

$W=0.18E=(0.18)(1.68\cdot 10^6)=3.02\cdot 10^5 J$

Learn more about power and work:

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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right

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Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

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Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

g: Gravitational Acceleration

θ: Angle with the vertical

N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

m*g*cos(θ) - 0 = m*vf^2 / R

g*cos(θ) = vf^2 / R

vf^2 = R*g*cos(θ)

vf^2 = 1.45*32.2*cos(34)

vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

ΔK.E = ΔP.E

0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))

vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

vi^2 = 22.744

vi = 4.77 ft/s

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3 years ago

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is

HACTEHA [7]

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

$mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m$

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

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3 years ago

A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.

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Answer:

t = 6.09 seconds

Explanation:

Given that,

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Hence, the time interval of the marble is 6.09 seconds.

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Answer:

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2 years ago

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

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Answer:

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Explanation:

Using as convention:

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The x- and y- components of the initial velocity of the jet can be written as

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While the components of the velocity of the wind are

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So the components of the resultant velocity of the jet are

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