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Studentka2010 [4]
3 years ago
15

What are the odds of nonsmokers developing cardiovascular disease when compared to smokers?

Physics
2 answers:
77julia77 [94]3 years ago
6 0
The probability ("odds") is smaller for nonsmokers when compared
to smokers.

If we had as much hard, reliable, repeatable, scientific evidence linking
cardiovascular disease with visits to the Empire State Building as we have
linking it with smoking, the Empire State Building would have been torn down
50 years ago.
insens350 [35]3 years ago
3 0

Answer choices are:

a. Nonsmokers are four times more likely to develop cardiovascular disease.

b. Nonsmokers are twice as likely to develop cardiovascular disease.

c. Nonsmokers and smokers are equally likely to develop cardiovascular disease.

d. Nonsmokers are half as likely to develop cardiovascular disease.

____________________________________________________________

Correct answer choice is:

d. Nonsmokers are half as likely to develop cardiovascular disease.

____________________________________________________________

Explanation:

The consequence of previous or existing smoking on specific CVD risk parameters such as blood pressure and cholesterol appears to be of clinically trivial importance in personalities aged 46 years. In additional terms, smoking appears to be beyond all an unconventional risk factor for CVD in the working-age people. Stopping smoking in working age may thus decrease estimated CVD risk approximately to the equivalent level as personages who have never smoked.

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At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 395 N/C. At a distance r2 fro
Ksenya-84 [330]

Answer:

r₂/r₁ = 1.82

Explanation:

The electric field due to a point charge, has the following expression:

E =\frac{k*q}{r^{2}}

For a distance r₁, the magnitude of the electric field is 395 N/C, so we can solve for r₁², as follows:

r₁² = \frac{k*q}{395 N/C} } (1)

For a distance r2, the magnitude of the electric field is 119 N/C, so we can solve for r₂², as follows:

r₂² = \frac{k*q}{119 N/C} }

We can find the quotient r₂/r₁, from (1) and (2):

r₂/r₁ =\sqrt{395/119} = 1.82

7 0
4 years ago
If you wished to warm 100 kg of water by 15 degrees celsius for your bath, how much heat would be required? (give your answer in
Anit [1.1K]
For the answer to the question above, 
<span>Q = amount of heat (kJ) </span>
<span>cp = specific heat capacity (kJ/kg.K) = 4.187 kJ/kgK </span>
<span>m = mass (kg) </span>
<span>dT = temperature difference between hot and cold side (K). Note: dt in °C = dt in Kelvin </span>

<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
<span>Q = 6,280.5 KJ = 6,280,500 J = 1,501,075.5 cal</span>
6 0
3 years ago
A 25.0kg girl pushes a 50.0kg boy with a force of 100.0N. What is the acceleration of the girl?
MA_775_DIABLO [31]

Answer:

im pretty sure it should be 50.0

6 0
3 years ago
a railroad tie weights 920 N and is 2.6 m long. How much force is required to: pick it up off the ground? lift one end and rotat
lukranit [14]

1) The minimum force needed is 920 N

2) The minimum force is 460 N

3) The minimum force is 598 N

Explanation:

1)

We can answer this part by simply looking at the forces involved. In fact, there are two forces acting on the railroad:

  • Its weight, W, acting downward
  • The force applied to lift it, F, upward

So the net force on the railroad is

\sum F = F - W

where

W = 920 N is the weight of the railroad

In order to lift the railroad, the net force must be upward, so

\sum F \geq 0

And therefore

F\geq W

which means that the minimum force needed is equal to the weight of the railroad, 920 N.

2)

In this case, we have to use the principle of equilibrium of moments.

In fact, when the railroad rotates uniformly (=constant angular speed) about its end, it means that the moment produced by the weight (acting in one direction) is equal to the moment produced by the force applied (acting in the other direction). Therefore, we can write:

W \frac{L}{2} = F L

where

W = 920 N is the weight

L = 2.6 m is the length of the railroad

F is the force applied

We wrote L/2 on the left of the equation because the weight acts at the center of mass of the railroad (located at the midpoint), while on the right it is L because the force F is applied at the end of the railroad.

Solving for F,

F=\frac{W}{2}=\frac{920}{2}=460 N

3)

This problem is similar to the previous part, however in this case, the force applied F is applied 0.6 m from the end, pivoting around the opposite end.

This means that the distance between the point of application of the force F and the pivot is

L' = L - 0.6

where

L = 2.6 m

Therefore the equation for the equilibrium of moments becomes

W\frac{L}{2}=F(L-0.6)

and substituting

W = 920 N

L = 2.6

We find the magnitude of F:

W\frac{2.6}{2}=F(2.6-0.6)\\1.3W = 2F\\F=\frac{1.3}{2}W=\frac{1.3}{2}(920)=598 N

Learn more about forces:

brainly.com/question/8459017

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4 0
3 years ago
Why are choir concerts better in performance halls than in gymnasiums?
dlinn [17]

Constructive interference in performance halls and the elimination of echoes are reasons why choir concerts better in performance halls than in gymnasiums.

<h3>What is constructive interference?</h3>

Constructive interference is a phenomenon which occurs when two waves travelling in same direction and which are in phase add up together to produce a wave of greater amplitude.

Constructive interference occurs in performance halls while destructive interference occurs in gymnasiums.

Also in performance halls, echoes are minimized due to the padded walls and the curtains while echoes which disturb choir concerts occur in gymnasiums.

Therefore, choir concerts better in performance halls than in gymnasiums because of constructive interference in performance halls and the elimination of echoes.

Learn more about constructive interference and echoes at: https://brainly.in/question/2378717

3 0
2 years ago
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