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seraphim [82]
3 years ago
13

A large manufacturing company releases chemicals into the air every day. The chemicals eventually make their way into local bodi

es of water, killing many of the organisms that live in these areas.
Which best describes this situation?

A.habitat destruction due to pollution
B.habitat destruction due to poaching
C.pollution due to poaching
D.poaching due to habitat destruction
Physics
1 answer:
garri49 [273]3 years ago
6 0
A. Habitat deconstruction due to pollution
You might be interested in
A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder
Stella [2.4K]

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

\terxt{time for compression}=\text{time for half revolution}

\terxt{time for compression}=\dfrac{1}{2}\times T

\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}

Put the value into the formula

\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60

\terxt{time for compression}=0.2\ sec

We need to calculate the rate of internal energy

Using first law of thermodynamics

U=Q-W

\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}

Put the value into the formula

\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)

\dfrac{\Delta U}{\Delta t}=6.8\ kW

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}

\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}

Put the value into the formula

\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}

\Delta\theta=32.48^{\circ}

Hence, The temperature change per compression stroke is 32.48°.

6 0
4 years ago
A dock worker applies a constant horizontal force of 81.0 N to a block of ice on a smooth horizontal floor. The frictional force
AnnZ [28]

Answer:

The correct answer is:

(a) 84.240 kg

(b) 24.038 m

Explanation:

The given values are:

Force,

F = 81.0 N

Distance,

S = 13.0 m

Time,

t = 5.20 s

As we know,

The acceleration of mass will be:

⇒  a=\frac{2S}{t^2}

On substituting the given values, we get

⇒     =\frac{2\times 13.0}{(5.20)^2}

⇒     =\frac{26}{27.04}

⇒     =0.961538 \ m/s^2

(a)

The mass of the block will be:

⇒  m=\frac{F}{a}

On substituting the given values, we get

⇒       =\frac{81.0}{0.961538}

⇒       =84.240 \ kg

(b)

The final velocity after a given time i.e.,

t = 5.00 s

⇒  v=at

On substituting the values, we get

⇒     =0.961538\times 5.00

⇒     =4.8076 \ m/s

In time, t = 5.00 s

The distance moved by the block will be:

⇒  d=vt

On putting the values, we get

⇒     =4.8076\times 5.00

⇒     =24.038 \ m

5 0
3 years ago
Nếu tăng khoảng cách giữa hai điểm lên 4 lần thì lực tương tác tĩnh điện giữa chúng sẽ
ohaa [14]

Answer:

điện giữa chúng s

Explanation:

8 0
3 years ago
You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is
katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0
3 years ago
Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction
gizmo_the_mogwai [7]
<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>
5 0
3 years ago
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