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katovenus [111]

3 years ago

11

22. Sweat glands are found in human skin. These glands release sweat to help the body cool during exercise. What would MOST LIKE

LY happen during exercise if a person's sweat glands were NOT functioning?
A. The body would become overheated.
B. More sweat would be released.
C. The upper layer of skin would become thicker.
D. More hair would grow on skin.

1 answer:

Bingel [31]3 years ago

7 0

Answer:

A.

Explanation:

all i know is that sweat keeps your body from overheating

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Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

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For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
$r_1+r_2=L km$
The second equation is going to the total force acting on the charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Let's solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now we have a quadratic equation with following parameter:
$a=2\\ b=2L\\ c=-L^2$
We know that two solutions are:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
We need a positive solution. When we plug in all the numbers we get:
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When does the fall equinox occur in the northern hemisphere

aivan3 [116]

Here is your answer

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HOPE IT IS USEFUL

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A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g

Sliva [168]

To solve for force, you need to get the product of mass and acceleration.
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

$a = \frac{vf-vi}{t}$

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.

So we can put it into our formula now:

$a = \frac{0m/s-50m/s}{20ms}$

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.

$20ms$ x $\frac{1s}{1000ms}$ = $\frac{20s}{1000} = 0.02s$

So your new time is 0.02s. Now we put this time into the formula:

$a = \frac{0m/s-50m/s}{0.02s}$
$a = \frac{-50m/s}{0.02s} = -2,500 m/ s^{2}$

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.

So now we have our acceleration. Now using this, we can get our force.

$F= ma$

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is $kg.m/ s^{2}$ but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.

$150g = \frac{1kg}{1,000g} = \frac{150kg}{1,000} = 0.150kg$

Our new mass is 0.150kg.

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 $m/ s^{2}$

Now that all our units match, we can put that into our formula:

$F= ma$
$F= (0.150kg)(-2,500m/s^{2})$
$F = -375kg.m/ s^{2} or -375N$

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.

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