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NARA [144]
3 years ago
14

Rank from most to least, the amount of lift on the following airplane wings: A. Area 1500 m2 with atmospheric pressure differenc

e of 2.9 N/m2. B. Area 900 m2 with atmospheric pressure difference of 4.7 N/m2. C. Area 500 m2 with atmospheric pressure difference of 8.8 N/m2.
Physics
1 answer:
AveGali [126]3 years ago
5 0

Answer:

C) >A)>B)

Explanation:

We know that force per unit area is known as pressure.But we have to understand that this force should be perpendicular to the area.

 P = F/A

P=Pressure

F= Force

A=Area

A)

Given that

ΔP=2.9 N/m²

A=1500 m²

F= ΔP x A

F= 2.9 x 1500

F=4350 N

B)

Given that

ΔP=4.7 N/m²

A=900 m²

F= ΔP x A

F= 4.7 x 900

F= 4230 N

C)

Given that

ΔP=8.8 N/m²

A=500 m²

F= ΔP x A

F= 8.8 x 500

F=4400 N

4400 N > 4350 N > 4230

C) >A)>B)

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There are 3 CO atoms
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A negative charge is moving across a room from south to north. A magnetic field runs from east to west. In what direction is the
melisa1 [442]

Answer:

Downwards

Explanation:

Using Fleming's right hand rule, where the middle finger represents the negative charge moving in the south to north direction, the fore finger represents the magnetic field which points in the east to west direction.

If we are to follow this orientation and place the thumb, middle finger and fore-finger at right angles to each other, the thumb represents the direction of the magnetic force on the charge.

In this case, the magnetic force points downwards after setting the other orientations of the magnetic field and the negative charge's direction.

3 0
3 years ago
A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou
zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius  = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}

E=6.88\times10^{7}\ N/C

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d).  For, r = 7.00 cm

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}

E=5.43\times10^{6}\ N/C

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0
3 years ago
Objects 1 and 2 attract each other with a electrostatic force of 36.0 units .If the charge of object 1 is doubled AND the charge
AfilCa [17]

Answer:

216 units

Explanation:

F1 = kQ1Q2/r^2 = 36 (Coulomb's law)

F2 = k(2Q1)(3Q2)/r^2

= 6[kQ1Q2/r^2]

= 6F1

= 6 x 36

= 216 units

6 0
3 years ago
An unknown liquid has a mass of 30.70 g and a volume of 52.3 mL. What is the density of the liquid?
kolbaska11 [484]

Answer:

de sity=mass÷volume.........

6 0
3 years ago
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