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3 years ago

7

AN AIR CRAFT HEADS NORTH AT 320KM/H RELATIVE TO THE WIND THE WIND VELOCITY IS 80KM/HFROMNORTH .FIND THE VELOCITY OF THE AIR RELA

TIVE TO THE GROUND

1 answer:

makkiz [27]3 years ago

5 0

Answer:

Explanation:

See attached file for answer.

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Will upvote!!!!

juin [17]

D. potential energy, because there is a bunch of water pent up, essentially stationary, waiting to roll down the steep mountain from the peak, so to say. if the dam were to be removed it would become kinetic.

4 0

3 years ago

Read 2 more answers

What is the gauge pressure in Pascals inside a honey droplet of a 0.1 cm diameter? Assume that air is surrounding this droplet a

Vikki [24]

Answer:

The gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

diameter of the honey droplet, D = 0.1 cm

radius of the honey droplet, R = 0.05 cm = 0.0005 m

surface tension of honey, γ = 0.052 N/m

Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure = P₁ - P₀ = 2 (2γ / r)

Where;

P₀ is the atmospheric pressure

Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

Gauge pressure = 416 Pa

Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa

5 0

4 years ago

Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30

beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

$J=-D\frac{\Delta C}{\Delta x}$

$\Delta C$ is the rate in concentration

$\Delta x$ is the rate in thickness

D is the diffusion coefficient, where,

$D= D_0 exp(\frac{Q_d}{RT})$

Replacing D in the first law,

$J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}$

clearing T,

$T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}$

Replacing our values

$T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}$

$T=-\frac{80000}{-138.09}$

$T=575.16K$

4 0

4 years ago

It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su

san4es73 [151]

Answer:

$EMF = 5.01 \times 10^{-4} Volts$

Explanation:

Here we know that the EMF induced in this Field is given as

$EMF = vBL$

here B = perpendicular component of magnetic field

v = speed of the bird

L = length of the wings

now we have

$B = 5\times 10^{-5} sin40$

$v = 13 m/s$

$L = 1.2 m$

now we have

$EMF = (13)(3.21 \times 10^{-5})(1.2)$

$EMF = 5.01 \times 10^{-4} Volts$

8 0

3 years ago

An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is giv

konstantin123 [22]

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

So, the percent uncertainty in the calculated value of g is 8.91 %

6 0

3 years ago