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Flura [38]
3 years ago
10

The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine

fails, how far can it glide in terms of distance measured along the ground?
Physics
1 answer:
andre [41]3 years ago
3 0

The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,

R = h (\frac{L}{D})_{max}

Replacing,

R = 5000ft (7.7)

R = 38500ft

Converting this units to miles.

R = 38500ft (\frac{1mile}{5280ft})

R = 7.2916miles

Therefore the glide in terms of distance measured along the ground is 7.2916miles

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A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.
Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

4 0
2 years ago
Why is it important to apply critical thinking to spam e-mail?
BARSIC [14]
Spam email<span> can contain malicious computer code and viruses,it can be an attempt to commit fraud,and it can be an attempt to get personal and financial.</span>
5 0
3 years ago
Read 2 more answers
How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns
Tasya [4]

Complete Question

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.

The output from the secondary coil is  12 V

Answer:

The value  is  N_s  =  21 \  turns

Explanation:

From the equation we are told that

   The input voltage is  V_{in}  = 120 \ V

   The number of turns of the primary coil is N_p =  210 \  turn

    The output from the secondary is V_o =  12V

From the transformer equation

   \frac{N_p}{V_{in}}  =\frac{N_s}{V_o}

Here N_s is the number of turns in the secondary coil

=> N_s  =  \frac{N_p}{V_{in}}  *  V_s

=>N_s  =  \frac{210}{120}  *  12

=>N_s  =  21 \  turns

4 0
3 years ago
How to atoms behave in non-magnetic items?
Anastaziya [24]

Answer:

By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.

Explanation:

5 0
2 years ago
How many joules of work are done on an object when a force of 10 N pushes it 5 m?
zhenek [66]

Answer:

option C

Explanation:

given,                            

Force on the object = 10 N

distance of push = 5 m

Work done = ?              

we know,              

work done is equal to Force into displacement.

W = F . s            

W = 10 x 5              

W = 50 J                

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

5 0
3 years ago
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