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3 years ago

How can pollution affect an ecosystem? Provide specific examples from the video.

Physics

1 answer:

guapka [62]3 years ago

5 0

Answer:

Ecosystems and Air Quality. Research shows that air pollution may effect ecosystems. Air pollutants such as sulfur may lead to excess amounts of acid in lakes and streams, and can damage trees and forest soils. ... Ozone damages tree leaves and negatively affects scenic vistas in protected natural areas

hope it helps ☺️✌️

You might be interested in

What is meant by infinite slew rate

Dennis_Churaev [7]

Am infinite slew rate means that the changes in the output voltage occur immensely when the input voltage changes.
Slew rate is measurement of the response of an operational amplifier. For an ideal operational amplifier, time delay is negligible. Hence it has an infinite slew rate.
In simpler terms it means that it can provide output voltage simultaneously with the input voltage changes.

Hope this helps :)

3 0

3 years ago

A horse running at 38 m/s reduces to 12 m/s in 10 seconds. What is the horse’s acceleration?

telo118 [61]

Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you

8 0

3 years ago

A child drops a ball. The ball hits the ground and bounces. The graph to the left shows the velocity-time graph for the ball fro

WINSTONCH [101]

Answer:

Consider the velocity-time graph attached below.

The velocity-time graph represents the acceleration of a body under a force.

We can see that is the graph that if a child release the ball above the ground at A, it hits the ground at B. Bounces back with a reaches the top again at C, and hits the ground again at D.

The slope of velocity time graph represents acceleration. From A to B, velocity in increasing constantly with respect to time, which means constant acceleration from A to B. AS velocity increase, momentum of the ball also increases, which results in the increase of Kinetic energy.

At B, the ball hits the ground, the velocity decreases, momentum decrease s, because kinetic energy is transferred from the ball to the ground, due to which the ball would not attain the same height after the bounce.

Then the velocity remains negative at C, which means that now the ball is moving in opposite direction till C. It reaches its new at height at C, which is not the same as that of A because of lost in Kinetic Energy, and fall again.

4 0

3 years ago

An automobile traveling 92.0 km/h has tires of 64.0 cm diameter. (a) What is the angular speed of the tires about their axles? (

Semenov [28]

Answer:

76.4035 m

Explanation:

r = Radius = 0.32 m

$\omega_f$ = Final angular velocity = 0

$\omega_i$ = Initial angular velocity = 92 km/h

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{\dfrac{92}{3.6}}{0.32}\\\Rightarrow \omega=79.861\ rad/s$

The angular speed of the tires about their axles is 79.861 rad/s.

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-79.861^2}{2\times 2\pi \times 38}\\\Rightarrow \alpha=-13.355\ rad/s^2$

The magnitude of acceleration is 13.355 m/s²

Distance is given by

$d=\theta r\\\Rightarrow d=38\times 2\pi\times 0.32\\\Rightarrow d=76.4035\ m$

The distance moved while slowing down is 76.4035 m

6 0

3 years ago

A large ant is standing on the middle of acircus tightrope that is stretched with tension Ts. The rope has mass per unit length

77julia77 [94]

Answer:

Explanation:

The ant , at some point of time will be standing on the crest of moving wave passing under its feet . At that position the downward acceleration of the ant will be equal to its weight . Hence it will feel weightlessness . So at that time

Acceleration = g

Acceleration at the top position = ω²A where ω is angular velocity and A is amplitude of vibration ( transverse ) of crest .

ω =2 π n where n is frequency of transverse vibration of particles. This angular frequency will be equal to angular frequency of wave travelling on the rope .

frequency of travelling wave

n.= $\frac{1}{\gamma}\sqrt{\frac{T}{\mu} }$

ω =2 π x $\frac{1}{\gamma}\sqrt{\frac{T}{\mu}$

ω² = 4π² x T/γ²μ}

For weightlessness

ω² A = g

A= g x γ²μ / 4π² T

7 0

3 years ago