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3 years ago

1. What could be the energy of a flying aeroplane? 2. Is work the same as power?

Physics

1 answer:

Doss [256]3 years ago

7 0

Explanation:

The total energy of an aircraft flying in the atmosphere can be calculated using equation 1. [2]

E = ½ m v2 + mgh

A Boeing 737-300 has a maximum takeoff weight of 5.65 × 104 kg, a cruise altitude of h = 10,195 m, and cruise speed of 221 m/sec. Inserting these numbers into the above equation, we obtain 7.03 GJ for the energy at cruise conditions. [3] However, the engines mounted onto the wings of the plane are required to provide additional energy per time, power, in order to keep the aircraft flying at a constant altitude and speed

Work is the energy needed to apply a force to move an object a particular distance, where force is parallel to the displacement. Power is the rate at which that work is done.

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4 years ago

Imagine your teacher asks you to design an experiment where you test the effect of temperature on the growth of a plant. You hav

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Answer:

The correct answer will be-

1. Dependent variable- The growth of plant in the form of height

2. Independent variable- different temperature

3. Constant variable- The amount of water, amount of sunlight, type of soil.

Explanation:

A Scientific experiment must include three types of variables which are: The independent, dependent and the constant variable.

1. Independent variable- The variable which can be modified or changed either on its own or manually. The variable directly influences the variable to be studied. In the given condition, the independent variable is the different temperature provided to the plants.

2. Dependent variable- The variable which is being studied in the experiment and directly influenced by the independent variable is the growth of the plant which is measured in the form of height.

3. Constant variable- The variable which is kept constant throughout the experiment and remains the same which could be the amount of water amount of sunlight and type of soil.

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3 years ago

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4 years ago

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$