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3 years ago

How do paved parking lots and roads with concrete or asphalt affect water flow on the land?

1 answer:

8 0

On the regular ground, water usually seeps in to the soil, which has a number of benefits for the wildlife in that area. However, with asphalt and concrete, there is no soft and absorbent soil to take in the water, so it just keeps flowing down to the lowest part of land it can. Additionally, some water runoff can carry fertilizer and other harmful chemicals with it in to the oceans and lakes it’s dumped in, which harms the ecosystem in them as well.

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If an astronaut throws an object in space, the object’s speed will _____

BigorU [14]

The object's speed will not change.

In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
$\sum F=ma$
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.

7 0

3 years ago

d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s

Harrizon [31]

Answer:

$a' = 0.35 m/s^2$

Explanation:

Let say the empty wagon has mass "M"

now by newton's II Law we will have

$F = Ma$

now it is given that empty wagon is pulled with acceleration 1.4 m/s/s

now we will have

$F = 1.4 M$

now a child of mass three times the mass of wagon is sitting on the empty wagon

so here we have

$F = (M + 3M) a$

$1.4 M = 4M a'$

so we have

$a' = 0.35 m/s^2$

8 0

3 years ago

Help my in science state of matter

Arada [10]

I would say the last option, since with an increase in temperature, water molecules will speed up.

7 0

2 years ago

Read 2 more answers

The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earthâ€™s surface.

aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = $\frac{1}{3}$ g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = $\frac{1}{3}$ m * g₁ = $\frac{1}{3}$ w₁

⇒Weight of probe on Mars = $\frac{1}{3}$ Weight of probe on earth

6 0

3 years ago

A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant

Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0

3 years ago