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3 years ago

11

3. High triglycerides increase the risk for which one of the following conditions? A. Osteoporosis B. Thyroid disease C. Heart d

isease D. Type II diabetes

1 answer:

Brut [27]3 years ago

5 0

The answer is most likely C. Heart disease

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in positive numbers less than 1, the zeros between the decimal point and non zero are _____ significant

Margaret [11]

Zeros the left of the decimal but still less than one are non significant but to the right they are.
For example: 0.003 (only 3 is significant) but for 0.030 (two are significant)

8 0

3 years ago

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105

Gnesinka [82]

Answer:

Q1 = 7.25*10^(-16) C

Explanation:

We are given;

electric field strength = (1 x 10^5 N/C

drag force (F) = 7.25 x 10^(-11) N

The question says it's moving with constant velocity. This means that he particle is in equilibrium and not accelerating.

Columbs law force of attraction or repulsion between two charges is given as;

F=(KQ1Q2)/r²

Now, electric field strength is given as the formula;(K*Q2)/r², thus plugging the relevant values gives us;

7.25 x 10^(-11) N= (1 x 10^(5) N/C)Q1 Q1 = 7.25 x 10^(-11) /(1 x 10^(5))

Q1 = 7.25*10^(-16) C

5 0

3 years ago

Read 2 more answers

Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. Find out the force ap

Alenkasestr [34]

Answer:

Force=10N

Velocity=13m/s

Explanation:

Given,

Mass = 5kg

initial velocity=3m/s

Final velocity=7m/s

Time=2s

Now acceleration= v-u/t

=7-3/2

=4/2

=2m/s²

So, acceleration=2m/s²

Now,

F=ma

=5*2

=10N

7 0

3 years ago

A common cylindrical copper wire used in a lab is 841 m long. Find the radius (in mm) of a wire necessary to have 0.5 Ohms of re

slava [35]

The relationship between the resistance R of a wire and its resistivity $\rho$ is given by
$R= \frac{\rho L}{A}$
where L is the length of the wire and A is its cross sectional area.

In the problem, we have $R=0.5 \Omega$ , $\rho = 1.68 \cdot 10^{-8} \Omega m$ and $L=841 m$ . So we can solve the find the area A:
$A= \frac{\rho L}{R}=2.83 \cdot 10^{-5} m^2$

For a cylindrical wire, the cross sectional area is given by
$A= \pi r^2$
where r is the radius. We know the value of the area A, so now we can find the radius of the wire:
$r= \sqrt{ \frac{A}{\pi} }= \sqrt{ \frac{2.83 \cdot 10^{-5}m^2}{\pi} }=0.003 m=3.0 mm$

3 0

3 years ago

What does the addition of two vectors give you?

kirill [66]

Answer:

D. Resultant Vector

Explanation:

By definition, adding 2 vectors gives a resultant vector

6 0

4 years ago