State two (2) examples of osmosis occurring in everyday life

Tina's calculations of the tarantula found that the spider was able to cover 20 centimeters in five seconds what was the average

Sergeu [11.5K]

<span>(20 cm)/(5 sec) = (0.20 meters)/(5 seconds)
</span>

8 0

3 years ago

What are some actions that can be taken to remedy air pollution? Select all that apply. Run your air conditioner at a higher set

Sati [7]

Answer: Government passing regulatory laws

Car emissions testing

Carpooling and less driving overall

Air pollution results when a mixture of particles and gases that can reach in harmful concentrations both outdoors as well as indoors. This increase in concentration can lead to rise in the temperatures and global warming it is also related with the health hazards. Soot, smoke, pollen, carbon dioxide, carbon monoxide and methane are the examples of the common pollutants.

Government passing regulatory laws: The government can ban some of the fossil fuels to prevent air pollution, and some vehicles which are likely to emit more partially combustible gases.

Car emission testing: Nowadays, cars and other vehicles are equipped with catalytic converters which converts the uncombustible gaseous products into less harmful gases which are emitted from the vehicle. The car gas emission testing should be performed to either allow a vehicle or ban the movement of the vehicle.

Carpooling and less driving overall: Carpooling is the sharing of one vehicle with more than one individual. This will prevent the need of others to drive another vehicle. The less the number of vehicles on the road the lesser will be the emission of gases and less pollution will be caused.

4 0

3 years ago

three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25

ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

GarryVolchara [31]

I'm in physics, but I think the answer is 22. The angle of reflection is reflecting whatever the angle was so, in this case, it must be 22

3 0

3 years ago

Read 2 more answers

A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$