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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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5.6 which would be acidic!
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10. Mario is making chocolate. In part of the chocolate-making process, he changed the phase from a liquid to a solid. What happ
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B

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Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the
IRISSAK [1]

Answer:

a₁ = 0.63 m/s²  (East)

a₂ = -1.18 m/s²  (West)

Explanation:

m₁ = 95 Kg

m₂ = 51 Kg

F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a   (⇒)

F = m₁*a₁         ⇒      60 N = 95 Kg*a₁    

⇒  a₁ = (60N / 95Kg) = 0.63 m/s²  (⇒)  East

To get the acceleration (magnitude and direction) of the woman we apply

∑Fx = m*a   (⇒)

F = -m₂*a₂         ⇒      60 N = -51 Kg*a₂    

⇒  a₂ = (60N / 51Kg) = -1.18 m/s²  (West)

For every case we apply Newton’s 3 d Law

8 0
3 years ago
What is the relationship between the applied force of a hanging mass on a spring and the spring force of the spring?
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Answer:

elastic force and weight are related to the acceleration of the System.

Explanation:

The relationship between these two forces can be found with Newton's second law.

        F_{e} - W = m a

        K x - m g = m a

We see that elastic force and weight are related to the acceleration of the System.

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7 0
3 years ago
Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp
In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the  final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M  = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0
3 years ago
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