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3 years ago

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Which of the following is a risk of obesity?

2 answers:

katrin2010 [14]3 years ago

7 0

Answer:

This question is incomplete but the completed is below

Which of the following is a risk of obesity?

A. Bone loss

B. Sleep apnea

C. Liver damage

D. Addiction

Explanation:

Obesity is a medical condition in which the body has accumulated excess body fats. This condition exposes the body to other medical complications/issues such as diabetes, atherosclerosis, cardiovascular diseases and in some forms of cancer.

Sleep apnae is a serious sleeping disorder that causes one to have an interrupted breathing pattern during sleep. Obesity can also cause this, as excess fat could disturb the windpipe or even the heart causing heavy and irregular breathing during sleep.

Kruka [31]3 years ago

6 0

What are the multiple choice answers?

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A car of mass 1535 kg collides head-on with a parked truck of mass 2000 kg. Spring mounted bumpers ensure that the collision is

dexar [7]

A car of mass 1535 kg collides head-on with a parked truck of mass 2000 kg. Spring mounted bumpers ensure that the collision is essentially elastic. If the velocity of the truck is 17 km/h (in the same direction as the car's initial velocity) after the collision, what was the initial speed of the car <u>20kmh</u>

<h3>What is collision ?</h3>

A collision in physics is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.

The following are a few instances of physical encounters that scientists might classify as collisions:

Legs of an insect are said to collide with a leaf when it falls on one.
Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.

To learn more about collision from the given link:

brainly.com/question/27736776

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7 0

1 year ago

Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.

babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

$d\sin \theta =m\lambda$

For Destructive interference

$d\sin \theta =(m+\frac{1}{2})\lambda$

where $\lambda =wavelength$

$d=slit\ width$

m=order of maxima and minima

for second order maxima i.e. $m=2$

For smallest separation taking $\lambda =400 nm, \theta =90^{\circ}$

$d\sin 90=2\times 400\times 10^{-9}$

$d=0.8\times 10^{-6}$

$d=0.8\mu m$

6 0

3 years ago

When does a lightbulb carry more current, (a) immediately after it is turned on and the glow of the metal filament is increasing

emmasim [6.3K]

A lightbulb carries more current immediately after it is turned on and the glow of the metal filament is increasing; option A.

<h3>What is current?</h3>

Current refers to the flow of electric charges typically electrons.

Current flowing through a metallic material decreases with increase in temperature of the material.

This is because the resistance of the metal increases with increase in temperature.

Therefore, for a light bulb, the current flow through it will be maximum when it is just turned on because the temperature, and hence the resistance of the filament is at its lowest.

In conclusion, current flow decreases with increase in resistance.

Learn more about current and resistance at: brainly.com/question/24858512

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8 0

1 year ago

A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric fi

Andrej [43]

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

4 0

3 years ago

A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (

Margarita [4]

Answer:

A) $h = 69.58 m$

D) $v = 58.12 \frac{m}{s}$ (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

$V_{ox} = V_{o}Cos\alpha _{o}$ Formula (1)

$V_{oy} = V_{o}Sin\alpha _{o}$ Formula (2)

Where:

Vo: Initial velocity in m/s

$\alpha_{o}$ : Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

$v_{y}^{2} = v_{oy}^{2} -2gy$ Formula (3)

Where:

$v_{f}^{2}$ : Final speed component in vertical direction in m/s

$v_{oy}^{2}$ : Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

$v_{x} = v_{ox}$ Because the movement is uniform in the x direction (constant speed)

$v_{y} = v_{oy}-g*t$ Formula (4) Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

$v_{o} = 65.2 \frac{m}{s}$

$\alpha _{o}$ = 34.5º above the horizontal

$g=9.8 \frac{m}{s^{2}}$

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

$v_{ox} = 65.2*Cos34.5 = 53.7\frac{m}{s}$

$v_{oy} = 65.2*Sin34.5 = 36.93\frac{m}{s}$

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

$0=(36.93)^2-2*9.8*h$

$h=\frac{ (36.93)^2}{2*9.8} = 69.58 m$

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

$v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}$

$v_{x} = v_{ox} = 53.7 \frac{m}{s}$

$v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}$ (Speed magnitude)

$\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) = tan^{-1} (\frac{22.23}{53.7})$

α = 22.49° (Speed direction above the horizontal)

6 0

3 years ago