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3 years ago

First sign of lung cancer

2 answers:

5 0

the most common first sign of lung cancer is a constant cough. there could also be coughing up blood or reddish colored spit, and chest pain when you laugh, cough, and breathe.

vodka [1.7K]3 years ago

3 0

So, my Nana had lung cancer and she just recently passed away from it because they could not find her a new lungs for a transplant. Signs aren't typically noticeable or seem severe until you get tested and realize you have lung cancer. Some signs would be a new cough that doesn't go away and maybe even have a very heavy chest, shortness of breath, or coughing up blood sputum (spit or phlegm) A lot of the "symptoms" could be similar to a cold or even asthma. I would say if you don't have asthma or asthma history it would be good to go to the doctor to get it checked out. Like I said they are all very similar to a cold or something like that but should be checked out if continuous.

I hope this helps :)

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Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1

luda_lava [24]

Answer:

a. 3.07 b. 1.26

Explanation:

Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j

Since A + B = -3.07i + 3.17j + b1i + b2j

= (-3.07 + b1)i + (3.17 + b2)j

So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j

Comparing components,

-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)

a. From (1), b1 = 3.07

b. From(2) b2 = 4.43 - 3.17 = 1.26

4 0

2 years ago

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

fiasKO [112]

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}$

h = 2.3 m

$v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s$

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

8 0

2 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

3 years ago

An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (

Afina-wow [57]

Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

this because we have an isosceles triangle, then the cathetus length is

hypotenuse/ $\sqrt{2}$

using Pythagoras, here the hypotenuse is 90, then the cathetus are of length

90/ $\sqrt{2}$ km/h= 63,6396 km/h.

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east, here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

$\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h$ ,

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= $sin^{-1}'frac{63,6396}{629,5851}$ =5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form