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Rama09 [41]
3 years ago
14

Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a

bove the horizontal, in 3.00 s. You can neglect the person's kinetic energy. Express your answer in horsepower
Physics
1 answer:
Viktor [21]3 years ago
3 0

Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, \theta=25^{\circ}

Time, t = 3 s

Let h is the height of the hill from the horizontal,

h=l\ sin\theta

h=12\times \ sin(25)

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

P=\dfrac{E}{t}

P=\dfrac{mgh}{t}

P=\dfrac{55.8\times 9.8\times 5.07}{3}

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

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Answer:E.

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8 0
3 years ago
2..Read the following statements carefully and choose the correct option
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statement 1 is correct

2 not sure

Explanation:

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A plane glides towards a ground-based radar dish. Radar locates the plane at a distance D = 22 km from the dish, at an angle θ =
vfiekz [6]

The correct answer is :

Unit vectors I and j along the x-axis and y-axis, respectively, define the Cartesian coordinate system. The radial unit vector r, which indicates the direction from the origin, and the unit vector t, which is orthogonal (perpendicular) to the radial direction, together create the polar coordinate system.

We can obtain the horizontal component by applying the trigonometric identity of Cos(Ф), and if we obtain the component on the x axle, such as 22000 (m)×Cos(51°) = x, we may determine that x = 13845.05 metres. We need to obtain the vector components because we already know the distance and the angle.

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8 0
1 year ago
A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a
olga nikolaevna [1]

Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

So

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

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Option - 4

5 0
3 years ago
You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of
alina1380 [7]

Answer:

The force constant ,I = 2394N/m

Explanation:

Given:

Weight of crate,Wg = 1470N

Theta = 22.0°

Kinetic friction,Fk= Fs(max) = 550N

Total length of ramp =8.0m

If y =0 at the bottom of the ramp

y1 = d Sin theta

y1 = 8 × Sin 22°

y1 = 3.0m

y2 = 0

V1=1.8m/s

V2 = 0

The equation combining the gravitational and elastic potential energy, and the work energy theorem is given by:

K1 + Ugrav1 + Uel1 + W = K2 + Uel1 ..eq1

Where KE is given by: 1/2mv^2

Gravitational potential energy is given by: Ugrav = mgy ...eq2

Elastic potential energy = Uel = 1/2Kx^2 ..eq3

The restoring force constant of a spring compressed by a distance x is given by:

Fx = Kx ..eq4

Workdone by a constant force is given by W = Fscostheta ...eq5

Where s = displacement

Theta = the angle between the force and the displacement

Work,W = Wf = 550 ×8 × cos 180°

W = -4400J

From the weight of the crate.mass is:

Wg/g = m

1470/9.8 = 150kg

K1 = 1/2 ×150×1.8^2 = 243m/s

The crate comes to rest at K2=0

Ugrav1 = 150 × 9.8 × 3 = 4410J

Ugrav2 = 150 × 9.8 x 0 = 0J

Uel1= 0 Spring at equilibrium

Substituting the values of the energies and work

253 + 4410 + 0 - 4400 = 0 + 0 + Uel1

Uel1 = 253J

Substituting into eq3

253 = 1/2 Kx^2 = 1/2 Kx(x)

Kx = 506/x

Since crate remains at rest,we use Newton's 2nd law

Fx = fs + Wsin theta

Fx = 550 + 1470 sin 22

Fx = 1100.7N

Substituting into eq4

Kx = 1100.7

X = 506/1100.7 = 0.46m

Kx = 1100.7

K = 1100.7/0.47

K = 2394N/m

5 0
3 years ago
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