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Svetllana [295]

3 years ago

7

What are some of the negative effects of ecotourism on an ecosystem?

1 answer:

oksano4ka [1.4K]3 years ago

4 0

Tons of stuff!
Footpath erosion, increased usage of travel vehicles, construction of hotels and other attractions, ect ect.

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Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel

krek1111 [17]

Answer:

The correct answer is <u>option (A) that is KEA > KEB .</u>

Explanation:

Let us calculate -

If the object is straighten up and inclined plane , the work done is

$W=F_d- F_f_r_id-F_gh$

$W=F_d-\mu_kmgdcos\theta-mgdsin\theta$

The change in kinetic energy is ,

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2$

At the top of the inclined plane , the velocity is zero

So,

$\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2$

$\Delta KE=-\frac{1}{2}m\nu_0^2$

From the work energy theorem , we have $W=-\Delta K$ in case of friction , so

$\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta$

$KE=Fd-\mu_kmgdcos\theta-mgdsin\theta$

For object A-

$KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta$

For object B

$KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta$

$KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)$

Thus , larger mass is going to mean less total work and a lower kinetic energy .

From the above results , we get

$KE_A >KE_B$

<u>Therefore , option A is correct .</u>

6 0

3 years ago

A 0.100-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

Gennadij [26K]

Answer:

Change in momentum of the stone is 3.673 kg.m/s.

Explanation:

Given:

Mass of the ball on the horizontal the surface, m = 0.10 kg

Velocity of the ball with which it hits the stone, v = 20 m/s

According to the question it rebounds with 70% of the initial kinetic energy.

We have to find the change in momentum i.e Δp

Before that:

We have to calculate the rebound velocity with which the object rebounds.

Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".

⇒ $KE_1=0.7\times \frac{mv^2}{2}$

⇒ $KE_1=0.7\times \frac{0.10\times (20)^2}{2}$

⇒ $KE_1=0.7\times \frac{0.10\times 400}{2}$

⇒ $KE_1=14$ Joules (J).

Rebound velocity "v1".

⇒ $KE_1=\frac{m(v_1)^2}{2}$

⇒ $v_1 = \sqrt{\frac{2KE_1}{m} }$

⇒ $v_1 = \sqrt{\frac{2\times 14}{0.10} }$

⇒ $v_1=16.73$

⇒ $v_1=-16.73$ m/s ...as it rebounds.

Change in momentum Δp.

⇒ $\triangle p= m\triangle v$

⇒ $\triangle p= 0.10\times (20-(-16.73)$

⇒ $\triangle p= 0.10\times (20+16.73)$

⇒ $\triangle p= 0.10\times (36.73)$

⇒ $\triangle p = 3.673$ Kg.m/s

The magnitude of the change in momentum of the stone is 3.673 kg.m/s.

5 0

3 years ago

Investigator Robbins is called to the scene of a robbery. The officer on the scene thinks that the suspect entered the home from

ICE Princess25 [194]

Answer:

3

Explanation:

Since the evidence found points to the window on investigation remember there was a foot print below the window as well as the window been opened.

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3 years ago

Use Kirchoff first law and second law to derise the expression for the total resistancs

mestny [16]

Answer:

Chupapi munyanyo

Explanation:

4 0

3 years ago

The temperature of 5 pounds of water is 40 degrees. Btus are added until the temp of

Vikentia [17]

Answer:

69becuz u smell l i k e c h e e z e

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3 years ago