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2 years ago

ASAP please!! The force vectors on an aircraft are as shown. Find the net (resultant force).

Physics

1 answer:

ioda2 years ago

6 0

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

$W_D=3600 N\cdot\sin 27^\circ=1634N$

acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

The magnitude of the net force is:

$|F|=\sqrt{5366^2+829^2}N = 5430N$

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The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

$I_2 = 4W/m^2$

Therefore the intensity at five times this distance from the source is $4W/m^2$

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3 years ago

If you stood on a planet having a mass four times higher than Earth's mass, and a radius two times 70) lon longer than Earth's r

BigorU [14]

CHECK COMPLETE QUESTION BELOW

you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?

A) four times more than you do on Earth.

B) two times less than you do on Earth.

C) the same as you do on Earth

D) two times more than you do on Earth

Answer:

OPTION C is correct

The same as you do on Earth

Explanation :

According to law of gravitation :

F=GMm/R^2......(a)

F= mg.....(b)

M= mass of earth

m = mass of the person

R = radius of the earth

From law of motion

Put equation b into equation a

mg=GMm/R^2

g=GMm/R^2

g=GM/R^2

We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have

m= 4M

r=(2R)^2=4R^2

g= G4M/4R^2

Then, 4in the denominator will cancel out the numerator we have

g= GM/R^2

Therefore, g remain the same

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3 years ago

The tin can with water in its bottom is heated to boil water and the steam is allowed to escape for some time. The open mouth is

Norma-Jean [14]

Explanation:

Water does expand with heat (and contract with cooling), but the amount of expansion is pretty small. So when you boil a can filled with water and seal it, the water will contract slightly as it cools. The can may kink slightly, but that will be it. Actually, most likely the only things you will be able to see is then top and bottom will be sucked in and go concave. Just like a commercial can of beans.

Now if you have a can with a little water and a big air space, things are completely different.

As the water boils, water vapour is given off. Steam. Let it boils for a minute just to make sure (nearly) all the air is expelled and the can is filled with steam.

Now when you put the lid on and cool the can, that steam condenses back to water, and goes from filling the can to a few drops of water. The can is now filled (if that is the right word) with a near vacuum, The air pressure, 15 lbs/square inch, will be pressing on every surface of the can, with nothing inside the can to resist it.

The can will crumple before your eyes.

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3 years ago

Suppose you have three identical metal spheres, AA, BB, and CC. Initially sphere AA carries a charge qq and the others are uncha

Sedaia [141]

Complete Question

Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?

a. 3/8q

b. 1/4q

c. 3/4q

d. q

e. 5/8q

f. 1/3q

g.1/2q

h. 0

Answer:

The correct option is b

Explanation:

From the question we are told that

The charge carried by A is q C

The charge carried by B is 0 C

The charge carried by C is 0 C

When A and B are brought close and then separated the charge carried by A and B is mathematically evaluated as

$\frac{ 0 + q}{2} = \frac{q}{2}$

When C and B are brought close and then separated the charge carried by C and B is mathematically evaluated as

$\frac{0 + \frac{q}{2} }{2} = \frac{q}{4}$

When C and A are brought close and then separated the charge carried by C and A is mathematically evaluated as

$\frac{\frac{q}{4} + \frac{q}{2} }{2} = \frac{3q}{8}$

Looking at these calculation we can see that the charge carried by B is

$\frac{q}{4} C$

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2 years ago

A number of conditions are required for a population to be in Hardy-Weinberg equilibrium. Which of the following are correct des

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Answer: A, B, F

Explanation:

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