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4 years ago

Changing your toilet can reduce water pollution because _______.

2 answers:

8 0

More efficient plumbing uses less water and creates smaller volume of waste is the answer why changing your toilet can reduce water pollution. If there is a problem in your toilet, you will use more water. The more water you pour, the more water pollution is added.

777dan777 [17]4 years ago

5 0

Answer:D

Explanation:

You might be interested in

List the characteristics properties of all waves. at what speed do electromagnetic waves travel in a vacuum

ad-work [718]

All electromagnetic waves travel at

299,792,458 meters per second

in vacuum.

7 0

4 years ago

What properties are in semiconductors

scoray [572]

Semiconductors are substances with properties somewhere between them. Conductors such as gold, silver and copper have low resistance and conduct electricity easily.

4 0

4 years ago

A rubber ball moving at a speed of 5 m/s hit a flat wall and returned to the thrower at 5 m/s. Which statement

Step2247 [10]

Answer:

d. Its magnitude and its direction both remained the same.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

This ultimately implies that, the law of conservation of momentum states that if objects exert forces only on each other, their total momentum is conserved.

In this scenario, a rubber ball moving at a speed of 5 m/s hit a flat wall and returned to the thrower at 5 m/s. Thus, the statement which correctly describes the momentum of the rubber ball is that its magnitude and its direction both remained the same because its velocity didn't change while returning to the thrower.

4 0

3 years ago

n astronaut who weighs 800 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration

ANTONII [103]

Answer: 0.29 kN

Explanation:

We have the following data:

$W_{E}=800 N$ is the weight of the astronaut on Earth

$g_{E}=9.8 m/s^{2}$ is the free fall acceleration due gravity on Earth (directed downwards)

$g_{Z}=3 m/s^{2}$ is the free fall acceleration due gravity on Zuton (directed downwards)

$a=0.5 m/s^{2}$ is the acceleration of the spaceship at litoff (directed upwards)

We have to find the <u>magnitude of the force</u> $F$ the space ship exerts on the astronaut.

Firstly, we have to know weight has a direct relation with the mass and the acceleration due gravity. In the case of Earth is:

$W_{E}=mg_{E}$ (1)

Where $m$ is the mass of the atronaut.

Isolating $m$ :

$m=\frac{W_{E}}{g_{E}}$ (2)

$m=\frac{800 N}{9.8 m/s^{2}}$ (3)

$m=81.63 kg$ (4)

Now that we know the mass of the astronaut, we can find its weight on Zuton:

$W_{Z}=mg_{Z}$ (5)

$W_{Z}=(81.63 kg)(3 m/s^{2})$ (6)

$W_{Z}=244.89 N$ (7)

Then, we can calculate the force the space ship exerts on the astronaut by the following equation:

$F-W_{Z}=m.a$ (8)

Isolating $F$ :

$F=m.a+W_{Z}$ (9)

$F=(81.63 kg)(0.5 m/s^{2})+244.89 N$ (10)

$F=285.7 N \frac{1 kN}{1000 N}=0.285 kN$ (11)

Finally:

$F=0.285 kN \approx 0.29 kN$

5 0

4 years ago

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

DENIUS [597]

Answer:

terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

Given the data in the question;

we know that, the force on a body due to gravity is;

$F_g$ = mg

where m is mass and g is acceleration due to gravity

Force of drag is;

$F_d$ = $\frac{1}{2}$ pCAv²

where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.

Terminal velocity is reach when the force of gravity is equal to the force of drag.

$F_g = F_d$

mg = $\frac{1}{2}$ pCAv²

we solve for v

v = √( 2mg / pCA )

so we substitute in our values

v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )

v = √( 1685.6 / 0.122015 )

v = √( 13814.6949 )

v = 117.54 m/s

v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr

Therefore terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

5 0

3 years ago