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KIM [24]
3 years ago
5

Gwen is researching the effects of a small asteroid impact in an area that has started to regrow after being hit ten years ago.

She considers the event to be a long-term change because the species of plants that were present had to start completely over from seeds after all of the plants in the area were killed by the impact. What is the error in Gwen’s thinking?
Physics
2 answers:
velikii [3]3 years ago
5 0

Answer:

D) This is a short-term change, not a long-term change.

Explanation:

Bumek [7]3 years ago
3 0

Answer:

Gwen’s assumption of asteroid hit as long term change is incorrect. Asteroid hit is not a long term change, instead, it is a short term change.  

Explanation:

Examples of short term changes are drought, flood, volcanic eruption, etc. A short term change occurs quickly and can immediately affect organisms but it doesn’t become a reason for species extinction. The effects of a short term change don’t prevail over a long span of time.

Examples of long term changes are ice age, global warming, deforestation, etc. Unlike a short term change, it takes time but the consequences are far-reaching. It can lead to species extinction.

In this question, asteroid hit is a quick and unexpected hazard, unlike the slow long term environmental changes.

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vivado [14]

Answer:

A ) Find the direct runoff depth; d =V A 22,698 _ 2.25x106 =0.01m =1cm The above result shows that the given hydrograph qualifies as a unit hydrograph. .

Explanation:

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3 years ago
A young man heaves a 6.5 kg rock at a velocity of 6.9 m/s. What is the kinetic energy of the rock?
earnstyle [38]

Answer:

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7 0
2 years ago
A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140
max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

u = 0\\\\v = 2.35\  \frac{m}{sec}\\\\d = 5.0 \ m\\\\

Using formula:

v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\

   = 0.55 \ \frac{m}{sec^2}\\\\

F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\

Calculating the Work by net force

W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\

The above work is converted into thermal energy.

Now,

F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\  by\ friction = -547.64 \ J

6 0
3 years ago
A swimmer is swimming to the left with a speed of 1.0 m/s when she starts to speed up with constant acceleration. The swimmer re
kolezko [41]

Answer:

Correct answer: t = 2.86 seconds

Explanation:

We first use this formula

V² - V₀² = 2 a d    

where V is the final velocity (speed), V₀ the initial velocity (speed),

a the acceleration and d the distance.

We will calculate the acceleration from this formula

a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10

a = 0.525 m/s²

then we use this formula

V = V₀ + a t  => t = (V - V₀) / a = (2.5 - 1) / 0.525 = 1.5 / 0.525 = 2.86 seconds

t = 2.86 seconds

God is with you!!!

6 0
3 years ago
The moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is
sleet_krkn [62]

Answer:

I = I₀ + M(L/2)²

Explanation:

Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.

The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.

The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀

The distance between the two axes is L/2 (total length of the rod divided by 2

From the parallel axis theorem we have

I = I₀ + M(L/2)²

5 0
3 years ago
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