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Charra [1.4K]
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.
Physics
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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A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c
Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

  • The plane wall is initially with constant properties is initially at a uniform temperature, To.
  • Suddenly the surface x=L is exposed to convection process such that T∞>To.
  • The other surface x=0 is maintained at To
  • Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

  1. There is only conduction in 1-D.
  2. The system bears constant properties.
  3. The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

T(x,0)=T_o

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u />T(0,t)=T_o<u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u />-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]<u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0
2 years ago
Which statement is true?
iogann1982 [59]
B 
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The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done =  mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = = \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})

kinetic energy = = \frac{1}{2}*4*(1.5^{2}-11^{2}) = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0
2 years ago
Why are force fields used to describe magnetic force?
Illusion [34]

Answer:

If I'm not working I think the answer is C.

7 0
2 years ago
A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
2 years ago
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