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AlexFokin [52]
2 years ago
7

How do volcanos effect ecosystems? in ur own words please!!

Physics
1 answer:
lana66690 [7]2 years ago
4 0

Answer:

well i guess i can try...

Explanation:

Volcanoes are a dangerous natural occurance that can greatly effect any ecosystems. There are various effects of volcanoes such as, lava flows, volcanic landslides, mud-flows, and ash flows, which generally result in total obliteration of any ecosystem in their path, they can also have long term beneficial effects to the environment.  One of these long term effects are, the creation of some of the most nutrisous, and fertile soil on earth, causing healthy plants to grow.

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Acceleration is defined as the change in velocity divided by
mihalych1998 [28]

Answer:

change in time

Explanation:

8 0
3 years ago
True or False
Allisa [31]
False?


science can be a motivation out of curiosity AND societal needs.

For example:battery operated  cars, are a curiosity and beneficial to the earths environment
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3 years ago
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A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0
3 years ago
A solid circular plate has a mass of 0.25 kg and a radius of 0.30 m. It starts rolling from rest at the top of a hill 10 m long
KATRIN_1 [288]

To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

The net height from the point where it begins to roll with an inclination of 30 degrees would be

h=Lsin30

h=10sin30

h=5m

In the case of Inertia would be given by

I = \frac{mR^2}{2}

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is

I = mk^2

\frac{mR^2}{2}= mk^2

\frac{k^2}{R^2}=\frac{1}{2}

Replacing in Energy conservation Equation we have that

Potential Energy = Kinetic Energy of Rolling Object

mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})

9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})

v^2 (1.5) = 98

v=8.0829m/s

Therefore the correct answer is C.

3 0
3 years ago
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg
katrin2010 [14]

Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

5 0
3 years ago
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