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2 years ago

6

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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Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

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The volume of the metal will be

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Density = 0.0158/0.00000225

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2 years ago

An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in

vesna_86 [32]

Answer:

height =20m

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1 year ago

Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V

Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

$Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter$

$Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter$

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

$R=\dfrac{\rho\times l}{A}$

Where,

R = Resistance

l = length

A = Area of cross section = πr²

$\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter$

Substituting the values we get

$R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}$

$R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}$

$R=\dfrac{1\times 10^{2}}{3.23}$

$R=30.95\ ohm$

Now by Ohm's Law,

$V= I\times R$

Substituting the values we get

$I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere$

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Current through Nichrome wire is 0.3879 Ampere.

4 0

1 year ago

Consider the static equilibrium diagram here. What is the angle F1 must make with the horizontal?

Nataly [62]

ANSWER

$\theta=35\degree$

EXPLANATION

Since the body is in equilibrium, total upward forces must equal total downward force.

Also the net horizontal forces acting on the body must be zero.

We need to resolve $F_1$ into vertical and horizontal components.

The horizontal component is

$x=F_1\cos\theta$ .

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$y=F_1\sin\theta$ .

Equating the up force to the downward forces gives,

$F_1\sin\theta + 20N=60N$ .

This implies that,

$F_1\sin\theta =60N-20N$ .

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Also the horizontal forces must be equal.

$F_1\cos\theta=57N...eqn2$ .

Dividing equation (1) by equation (2) gives,

$\frac{F_1\sin\theta}{F_1\cos\theta}=\frac{40}{57}$ .

$\Rightarrow \tan\theta=0.70175$

$\Rightarrow \theta=tan^{-1}(0.70175)$

$\Rightarrow \theta=35.0594$ .

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2 years ago

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Dafna1 [17]

Answer:

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Part A. Initially they ask us to find the moment of the whole system

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Part B.

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Part C.

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m a = [kg] [m / s2] = [N]

It is equivalent to force

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Acceleration is due to a net force applied

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The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1 f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0

1 year ago