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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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Differentiate between sound waves and seismic waves?

algol13

The only real difference is that common seismic waves travel through the ground and sound waves travel through the air. If you had a pipe attached to granite and you were listening to it, you might detect both.

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3 years ago

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The amount of light lost within a fiber-optic system is known as

densk [106]

Attenuation is the correct answer.

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3 years ago

17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What

Lubov Fominskaja [6]

Answer:

$\frac{D}{d} = 4.12$

Explanation:

As we know that resistance of one copper wire is given as

$r = \rho \frac{L}{a}$

here we know that

$a = \pi (\frac{d}{2})^2$

now we have

$r = \rho \frac{L}{\pi (\frac{d^2}{4})}$

$r = \rho \frac{4L}{\pi d^2}$

now we know that such 17 resistors are connected in parallel so we have

$R = \frac{r}{17}$

$R = \rho \frac{4L}{17 \pi d^2}$

Now if a single copper wire has same resistance then its diameter is D and it is given as

$R = \rho \frac{4L}{\pi D^2}$

now from above two equations we have

$\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}$

$D^2 = 17 d^2$

now we have

$\frac{D}{d} = 4.12$

3 0

3 years ago

1. A rock of granite has a mass of 50 kg. if it’s weight in water

igor_vitrenko [27]

Answer:

The first part of the question is asking about BUOYANT FORCE or UPTHRUST.

Upthrust =TRUE WEIGHT-APPARENT WEIGHT

TRUE WEIGHT=mg

TRUE weight=50kg×10m/s²

=500N

upthrust=500N-380N

FB=120N

volume of the rock=mass/density.

since the granite is completely submerged, the volume of the displaced liquid will be equal to the volume of the body.

upthrust=Vdg

120N=V×1000kg/m³×10m/s²

120N=V×10000kg/m²s²

120/10000=V

v=0.012m³

please mark brainliest, hope it helped

6 0

3 years ago

An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then

gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

$a=\frac{dv}{dt}=1.76$

So, the velocity can be obtained by integrating this expression:

$v=1.76t$

The velocity is, by definition: $v=\frac{dx}{dt}$ , so

$dx=1.76tdt\\x=1.76\frac{t^{2}}{2}$ .

Do x=11 in order to find the time spent.

$11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s$

At this time the velocity is: $v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}$

This velocity remains constant in the section 2, so for that section the movement equation is:

$x=v*t\\t=\frac{x}{v}$

The left distance is 89 meters, and the velocity is $6.2225\frac{m}{s}$ , so:

$t=\frac{89}{6.2225}=14.303s$

So, the total time is 14.303+3.5355s=17.838s

7 0

3 years ago