What are some methods humans have used to control flooding and what has been done to worsen the effects of flooding?

If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, what is the airplane's final velocity?

givi [52]

If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, the final velocity can be calculated using simple vector addition. In this case, the planes velocity is positive (+330 km/h) and head wind has a negative component (-18.0 km/h). Vector addition yields +330 km / h + (-18.0 km /h) = 312 km / h.

8 0

3 years ago

Why, on a sunny day, it is normally hot inside a greenhouse

Dahasolnce [82]

Because of the greenhouse gases inside the greenhouse, and the gases trap the heat from the sun so the plant don't freeze, hope this helps

5 0

3 years ago

An example of an unbalanced force is what

Nadya [2.5K]

Answer:

Any two forces acting on each other, which are not equal and opposite ( thereby causing motion ) are unbalanced forces.

Explanation:

Expierience

4 0

3 years ago

A sound has a sound level of 30 dB. Its intensity is what multiple of the standard reference level for intensities?

arsen [322]

<h2>Answer:</h2>

1000th multiple of the standard reference level for intensities.

<h2>Explanation:</h2>

The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;

β = 10 log (I / I₀) --------------------(i)

Where;

I₀ = reference intensity

Given from the question;

β = sound level = 30dB

Substitute this value into equation (i) as follows;

30 = 10 log (I / I₀)

Divide both sides by 3;

3 = log (I / I₀)

Take antilog of both sides;

10^(3) = (I / I₀)

1000 = I / I₀

Solve for I;

I = 1000I₀

Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)

7 0

3 years ago

A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have

denpristay [2]

1) The total mechanical energy of the rock is:
$E=U+K$
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
$E_i=U=mgh$
where $m=4 kg$ is the mass, $g=9.81 m/s^2$ is the gravitational acceleration and $h=5 m$ is the height.
Putting the numbers in, we find the potential energy
$U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J$

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
$E_f=K= \frac{1}{2}mv^2$
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
$K=U=196.2 J$

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
$W=196.2 J-0 J=196.2 J$

6 0

3 years ago