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3 years ago

Human activities can have an impact on natural disasters.

2 answers:

6 0

I think you wanted to know whether the statement in question is true or false. Assuming this fact, i am answering the question and hope that it helps you. It is a true fact that human activities can have an impact on natural disasters. The increase in global temperature due to emission of greenhouse gases can cause the sea level to rise and create hurricanes.

alukav5142 [94]3 years ago

4 0

By definition, we have to:

The environmental impact are the consequences that some of the human activities produce on the environment.

The concept can be extended to the effects of a catastrophic natural phenomenon.

Human activities, such as fuel combustion and deforestation, have intensified the greenhouse effect, causing global warming.

It has been estimated that if emissions of greenhouse gases continue at the current rate, the temperature of the planet could exceed historical values in 2047, with harmful effects on ecosystems, biodiversity and endanger the livelihood of people on the planet.

Answer:

True.

Human activities can have an impact on natural disasters.

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Using information about natural laws, explain why some car crashes produce minor injuries and others produce catastrophic injuri

Pani-rosa [81]

The Speed (Velocity) at which the car is moving.

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3 years ago

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Please help!<br> What is momentum?

VARVARA [1.3K]

Answer:

a property of a moving body that the body has by virtue of its mass and motion and that is equal to the product of the body's mass and velocity

Explanation:

Formula:

Momentum: Mass x Velocity

Hope it helps!

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3 years ago

The velocity of the transverse waves produced by an earthquake is 4.05 km/s, while that of the longitudinal waves is 7.695 km/s.

lora16 [44]

Answer:

x = 600 Km

Explanation:

given,

speed of transverse wave = 4.05 Km/s

speed of longitudinal wave = 7.695 Km/s

difference of the time of arrival of wave = 69.9 s

now,

t₂ - t₁ = 69.9 s

let x be the distance

$\dfrac{x}{4.05} - \dfrac{x}{7.695} = 69.9$

$x = \dfrac{69.9\times 4.05 \times 7.695}{7.695-4.05}$

x = 597.645 Km

x = 600 Km

distance of recorder from the site is 600 Km approximately.

4 0

4 years ago

__________ were initially developed to destroy the part of the brain that controls emotions.

sasho [114]

It’s Lobotomies on edg.

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3 years ago

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In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0

3 years ago