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stiks02 [169]
2 years ago
5

An airplane is heading due east. The airspeed indicator shows that the plane is moving at a speed of 370 km/h relative to the ai

r. If the wind is blowing from the north at 92.5 km/h, the velocity of the airplane relative to the ground is?
Physics
1 answer:
PolarNik [594]2 years ago
5 0

Answer:

Vp = 358.3m/s

Explanation:

Taking

North: UP

South: DOWN

East: RIGHT

West: LEFT

In the attachment, the velocity vector triangle is shown for the scenario. the triangle formed is a right angled triangle.

We can apply the Pythagora's Theorem.

370 ^ 2 = 92.5^2 + Vp^2    (Vp: plane velocity relative to ground)

Vp = sqr.rt(370^2 - 92.5^2)

Vp = 358.3m/s

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if a toaster transfers 100 joules of energy every ten seconds, what is the power rating of the toaster include the units in your
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Answer:

that is the solution to the question

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A square loop of wire has a perimeter of 4.00 mm and is oriented such that two of its parallel sides form a 13.0 ∘∘ angle with t
vivado [14]

Answer:

a) (2.436 × 10⁻⁷) Wb

b) (7.308 × 10⁻⁷) Wb

Explanation:

Magnetic flux is the dot product of the magnetic field vector and the Area vector.

Mathematically, it is given as

Φ = BA cos θ

where B = magnetic field strength

A = Cross sectional Area of the loop enclosed

θ = angle between the magnetic field and the plane of the area.

a) B = 0.250 T

To find A, the perimeter of the loop is given as 4.00 mm.

Perimeter of a square = 4L

4L = 4.00

L = 1.00 mm = 0.001 m

The area is given as L²

A = (0.001)² = 0.000001 m²

θ = 13°

Φ = BA cos θ

Φ = 0.25 × 0.000001 × cos 13°

Φ = 0.0000002436 Wb = (2.436 × 10⁻⁷) Wb

b) Another loop lies in the same plane but has an irregular shape, resembling a starfish. Its area is three times greater than that of the square loop

Φ = BA cos θ

B = 0.25 T

A = 3 × 0.000001 = 0.000003 m²

θ = 13°

Φ = 0.25 × 0.000003 × cos 13°

Φ = 0.0000007308 Wb = (7.308 × 10⁻⁷) Wb

Hope this Helps!!!

8 0
2 years ago
A highly charged piece of metal (with uniform potential throughout) tends to spark at places where the radius of curvature is sm
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Answer:

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Explanation:

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2 years ago
A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l
artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, \lambda_1=8\times 10^{-6}\ C

linear charge density of the other charge distribution, \lambda_2=-6\times 10^{-6}\ C

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

E=\frac{\lambda}{2\pi.r.\epsilon_0}

where:

\lambda= linear charge density

r = radial distance from the center of wire

\epsilon_0= permittivity of free space

Therefore,

E_1=E_2

\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}

\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}

\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}

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∴at y=6.29 cm the charge of the two distribution will be equal.

9 0
2 years ago
The graph indicates Linda’s walk.
Sedaia [141]
I think the right answer is the first one. If she stops moving her Position does not change any more-and the Graph Shows that after 6 seconds she stays at the Position of 5 m. If she Went Back to the start point the Graph would have Developed Back to 0m(decreased).
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2 years ago
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