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Mazyrski [523]
3 years ago
8

find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed

Physics
1 answer:
Cloud [144]3 years ago
6 0

Answer:

U=5*10^7ft.Ib

Explanation:

From the question we are told that

Weight  W= 10000bs

Altitude H=5000ft

Speed    V=125kts\\1kts=0.514m/s\\V=125*0.514=>64.25m/s

Generally the equation for Potential energy ids mathematically given as

Potential\ Energy\ U=mgh

U=Wh

U=10000*5000

U=5*10^7ft.Ib

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Which of the following describes a condition in which an individual would not hear an echo?
IrinaVladis [17]
If the echo (the reflected sound) reaches your ear less than about
0.1 second after the original sound, your brain doesn't separate them,
and you're not aware of the echo even though it's there.

If the echo comes from, say, a wall, 0.1 second means you'd have to be
about  17 meters away from the wall.  If you're closer than that, then the
echo reaches you in less than 0.1 second and you're not aware of it.

A. 30 meters . . .
     No.  You hear that echo easily

B.  you're standing within range of both sounds . . .
     No. You hear that echo easily, if you're at least 17 meters from the wall.

C.  less than 0.1 second later . . .
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D.  21.5 meters
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4 0
3 years ago
Read 2 more answers
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef
bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0
3 years ago
The radius of Venus (from the center to just above the atmosphere) is 6050 km (6050✕103 m), and its mass is 4.9✕1024 kg. An obje
Alecsey [184]

Answer:

(a) The initial speed required is 13116 m/s

(b) The escape speed is 10394 m/s

This problem involves the application of newtons laws of gravitation. The forces in action here are conservative and as a result mechanical energy is conserved.

The full calculation can be found in the attachment below.

Explanation:

In both parts (a) and (b) the energy conservation equation were used. Assumption was made that when the object is very far from the planet the distance from the planet's center approaches infinity and the gravitational potential energy approaches zero.

The calculation can be found below.

6 0
3 years ago
A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w
inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is v=\frac{d}{t}, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is d=v\cdot t

Now, the distance equation for the man would be:

d_{man}=v_{man}\cdot t=1.6\cdot t

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

d_{man}=d_{dog}

1.6\cdot t=3\cdot (t-1.8)

1.6\cdot t=3\cdot t-5.4

1.4\cdot t=5.4

t=\frac{5.4}{1.4}

t=3.8571s

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

d_{man}=1.6\cdot t=1.6\cdot (3.8571)

d_{man}=6.1714m

Finally, the dog catches up with the man 6.1714m later.

6 0
3 years ago
A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 222 kg and it
inn [45]

Answer:

The buoyant force is 3778.8 N in upward.

Explanation:

Given that,

Mass of balloon = 222 Kg

Volume = 328 m³

Density of air = 1.20 kg/m³

Density of helium = 0.179 kg/m³

We need to calculate the buoyant force acting

Using formula of buoyant force

F_{b}=\rho_{air}\times V_{b}\times g

Where, \rho_{air} = density of air

V = Volume of balloon

g = acceleration due to gravity

Put the value into the formula

F_{b}=1.20\times321\times9.81

F_{b}=3778.8\ N

This buoyant force is in upward direction.

Hence, The buoyant force is 3778.8 N in upward.

4 0
3 years ago
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