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2 years ago

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Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha

t average power must the winch supply in order to accelerate a 156-kg ultralight glider from rest to 24.9 m/s over a horizontal distance of 58.0 m? Assume that friction and air resistance are negligible, and that the tension in the winch cable is constant.

1 answer:

aliina [53]2 years ago

6 0

Answer:

P=627.47W

Explanation:

To solve this problem we have to take into account, that the work done by the winch is

$W=Fh$

the force, at least must equal the gravitational force

$F=Mg=(156kg)(9.8\frac{m}{s^2})=1258.8N$

with force the tension in the cable makes the winch go up.

The work done is

$W=(1258.8N)(58.0m)=73010.4J$

To calculate the power we need to know what is the time t. But first we have to compute the acceleration

The acceleration will be

$v_f^2=v_0+2ah\\a=\frac{v_f^2}{2h}=\frac{(24.9\frac{m}{s})}{2(58.0m)}=0.214\frac{m}{s^2}$

and the time t

$v_f=v_0+at\\t=\frac{v_f}{a}=116.35s$

The power will be

$P=\frac{W}{t}=\frac{73010.4J}{116.35s}=627.47W$

HOPE THIS HELPS!!

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nataly862011 [7]

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The reason the above values are correct are as follows:

The given parameters are;

Volume of the bottle, V = Half litre

Mass of the bottle, $m_b$ = 80 g

The volume of liquid in the bottle when filled, V = 500 cm³

The density of the olive oil with which the bottle is filled, ρ = 0.9 g/cm³

a. Required:

To calculate the mass of oil in the bottle

Solution:

The volume of oil in the bottle when the bottle is filled, V = 500 cm³

$Density, \ \rho = \dfrac{Mass}{Volume} = \dfrac{m}{V}$

The mass of the liquid, m = ρ × V

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b. The mass of the oil in the bottle, m = grams

The mass of the full bottle, $m_{filled}$ = m + $m_b$

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The mass of the full bottle, $m_{filled}$ = <u>530 grams</u>

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2 years ago

A 125g steel ball with a kinetic energy of .25j rolls along a horizontal track how high up an inclined track will the ball roll

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

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- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

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We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

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$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

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At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

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$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

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Since the book is at rest, these two forces are equal to each other and according to Newton's Third Law the reaction force to the force of gravity is equal but opposite to the weight of the book. This reaction force is the one that holds the book up on the shelf.

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Hello there.

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