All categories

2 years ago

Which type of environmental science career involves collecting information about how human events impact the environment?

Physics

2 answers:

Bas_tet [7]2 years ago

6 0

Answer: Environmental engineer

Explanation:

Environmental engineering can be defined as the area where impact of human activities on environment takes place.

Human activities as mining, deforestation, building dams has huge impact on the environment.

This is studied under environmental engineering. This subject also allows to find solution for the problems associated with the impact of human activities on environment.

SVEN [57.7K]2 years ago

4 0

C. environmental engineer

You might be interested in

State Pascal's principle of pressure . please help due tomorrow

Romashka [77]

Answer:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.

Explanation:

The pressure at any point in the fluid is equal in all directions.

3 0

3 years ago

What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d

Butoxors [25]

Calculate the pressure due to sea water as density*depth.
That is,
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is 101.3 kPa
Total pressure is 94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx

3 0

3 years ago

How can liquid pressure phenomena be used in daily life. Give one example with explanation of working?

N76 [4]

Answer:

An example in which liquid pressure phenomena can be used in daily life is in Water blasting

Explanation:

Water blasting refers application of pressurized water to remove materials from the surface of objects.

There are different varieties of water blasting, including;

Hydrocleaning; Cleaning enabled by the use of high pressure water

Hydrodemolition; Demolition or removal of concrete using pressurized water

Hydrojetting; The spraying of water under pressure on surfaces in order to remove surface contaminants.

8 0

3 years ago

A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___

34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8· $\hat j$ , in vector form

2) 1.0 km East = 1.0· $\hat i$ , in vector form

3) 1.6 km South = -1.6· $\hat j$ , in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8· $\hat j$ + 1.0· $\hat i$ +(-1.6· $\hat j$ ) = 1.2· $\hat j$ + 1.0· $\hat i$

Final displacement, = 1.0· $\hat i$ + 1.2· $\hat j$

Where;

Δx = The displacement in the x-direction = 1.0· $\hat i$

Δy = The displacement in the y-direction = 1.2· $\hat j$

The magnitude of the resultant displacement vector is given as follows

$\left | d \right |$ = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0

3 years ago

Compare the catching of two different water balloons.

Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = $F_{average}$ × Δt = mΔV

∴ $F_{average}$ = m·ΔV/Δt

∴ For Case A $F_{average}$ = 149.55×8/Δt = 1196.4/Δt N

For Case B $F_{average}$ = 598.2×8/Δt = 4785.6/Δt

Where Δt is the same for Case A and Case B, $F_{average}$ for Case B >> $F_{average}$ for Case B

Therefore, Case B involves the greatest force.

4 0

2 years ago