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Harman [31]
3 years ago
15

As a result of recent weather activity, there’s less water available for human consumption. Which biome is affected the most by

this change?
a)tundra
b)taiga
c)saltwater
d)freshwater
Physics
2 answers:
andreev551 [17]3 years ago
7 0
The correct answer for this question is this one: "d)freshwater"
<span>As a result of recent weather activity, there’s less water available for human consumption. The biome that is affected the most by this change is that <u>freshwater</u></span>
Hope this helps answer your question and have a nice day ahead.
Molodets [167]3 years ago
5 0

Answer: d)freshwater

The freshwater biome will be most likely to be affected by the weather conditions as the freshwater reservoirs accumulates water from sources like rain, flood and melting of snow. The reservoirs form are groundwater, river, lakes, ponds and others. In the absence of rain the water will not be accumulated in the biome and will effect the life forms living there.

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Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

• The net force in the perpendicular direction is

∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0

Solving the second equation for <em>n</em> gives

<em>n</em> = <em>mg</em> cos(21°)

<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)

<em>n</em> ≈ 1.83 N

Then the magnitude of friction is

<em>f</em> = <em>µn</em>

<em>f</em> = 0.25 (1.83 N)

<em>f</em> ≈ 0.457 N

Solve for the acceleration <em>a</em> :

-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

<em>a</em> ≈ -5.80 m/s²

so the block is decelerating with magnitude

<em>a</em> = 5.80 m/s²

down the ramp.

5 0
3 years ago
Roaring of a lion is different from the sound of the mosquito why?​
Soloha48 [4]

Answer:

The roaring of the lion is louder and has low pitch than the sound produced by a mosquito. Hence, they sound so different

4 0
3 years ago
A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f
Setler [38]

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

W_f = F_fs = 15*0.5 = 7.5 J

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

E_p = mgh = 0.1*10*1.5 = 1.5 J

Therefore, as elastic energy is converted to potential energy and work of friction:

E_e = W_f + E_p

kx^2/2 = 7.5 + 1.5 = 9 J

k = 9*2/x^2 = 18/0.05^2 = 7200 N/m

6 0
3 years ago
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0
4 years ago
An amoeba has 1.00×1016 protons and a net charge of 0.300 pC. (a) How many fewer electrons are there than protons? (b) If you pa
ladessa [460]

Explanation:

(a)   It is known that charge on a proton is equal to 1.6 \times 10^{-19} C. And, net charge is given as 0.3 pC which is also equal to 0.3 \times 10^{-12}C.

Therefore, we will calculate the number of electrons as follows.

           \frac{0.3 \times 10^{-12}C}{1.6 \times 10^{-19}C}

            = 1.87 \times 10^{6}

Hence, there are 1.87 \times 10^{6} fewer electrons are there than protons.

(b)   Now, we will calculate the fraction of protons that would have no electrons as follows.

               \frac{1.88 \times 10^{6}}{1 \times 10^{16}}

               = 1.88 \times 10^{-10}

Therefore, fraction of the protons that would have no electrons is 1.88 \times 10^{-10}.

3 0
3 years ago
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