What does not contain a lens? <br>A.binoculars<br>B.a mirror<br>C. an eye<br>D. a camera

Fill in the blank the process of evaporative cooling helps the body .

Nina [5.8K]

Answer:

The process of evaporative cooling helps the body is explained below in complete details.

Explanation:

Your body performs the application of the evaporative process when secreting. Sweat, which contains 90 percent water, commences evaporating. ... This appears in a cooling impression (described as evaporative cooling) that serves to sustain body temperature and cools the body down when it becomes too hot.

3 0

3 years ago

How to calculate the slope of position time graph?

Mashcka [7]

Answer:

The slope of a position-time graph can be calculated as:

$m=\frac{\Delta y}{\Delta x}$

where

$\Delta y$ is the increment in the y-variable

$\Delta x$ is the increment in the x-variable

We can verify that the slope of this graph is actually equal to the velocity. In fact:

$\Delta y$ corresponds to the change in position, so it is the displacement, $\Delta s$

$\Delta x$ corresponds to the change in time $\Delta t$ , so the time interval

Therefore the slope of the graph is equal to

$m=\frac{\Delta s}{\Delta t}$

which corresponds to the definition of velocity.

3 0

4 years ago

What standard of measurement do most countries use and discuss why such a standard is beneficial

frosja888 [35]

I mean i guess its the metric system?

6 0

3 years ago

The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car

Alex73 [517]

Answer:

15.8640053791 s

392.780107582 m

29.5184032275 m/s

Explanation:

0 denotes initial

x denotes displacement

c denotes car

t denotes truck

r denotes rear

$x_0_{cr}=-49.5\ m$

$a_c=0.6\ m/s^2$

$x_0_{t}=0$

$v_0_{c}=v_0_{t}$

For the car

$x_c=x_0+v_0_{c}t+\dfrac{1}{2}at^2$

The displacement of the truck will be

$x_t=v_tt$

From the above two equations we get

$x_c-x_t=x_0+v_0_{c}+\dfrac{1}{2}at-v_{t}t=26\\\Rightarrow 26=x_0+\dfrac{1}{2}at^2\\\Rightarrow 26+49.5=\dfrac{1}{2}0.6t^2\\\Rightarrow t=\sqrt{\dfrac{2(26+49.5)}{0.6}}\\\Rightarrow t=15.8640053791\ s$

The time taken is 15.8640053791 s

$x-x_0=v_{0}_{c}t+\dfrac{1}{2}at^2\\\Rightarrow x-x_0=20\times 15.8640053791+\dfrac{1}{2}0.6\times 15.8640053791^2\\\Rightarrow x-x_0=392.780107582\ m$