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3 years ago

How have increased carbon dioxide levels and temperatures affected living organisms?

Physics

2 answers:

balu736 [363]3 years ago

7 0

It leads to increase in temperature of the earth and it harms living being

Yuri [45]3 years ago

3 0

Answer:

it leads to increase in temperature of the earth and it harms living being

Explanation:

carbon dioxide is lite gas ,so it form a layer of carbon dioxide in the atmosphere.By our dairy activities such as driving ,burning etc release carbon dioxide which increase the layer of the carbon dioxide.the carbon dioxide layer absorb the heat energy from the sun so the temperature will increase and it will make global warming.the living organisms in the earth can not sustain the hot temperature hence it affect the living organisms in the earth

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a paper airplane gliding down towards the ground will experience the force of air resistance pushing up. the weight of the paper

Oduvanchick [21]

The net force acting on the airplane is 25N.

Forces acting on the paper airplane when it is in the air:

The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.

Given.

Weight of the paper airplane, F1 = 16N

The force of air resistance, F2 = 9N

Net force = F1 + F2

Net force = 25N

Thus, the net force acting on the airplane is 25N.

Learn more about the net force here:

brainly.com/question/18109210

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3 0

1 year ago

30 PTS

tigry1 [53]

Spreads out in the medium, first choice

7 0

3 years ago

A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr

mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

$f' = (\frac{v}{v\pm v_s})f$

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

$v_s$ is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

$v_s = -40.4 m/s$

So the frequency heard by the observer on the platform is

$f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz$

When the train has passed the platform, we have

$v_s = +40.4 m/s$

So the frequency heard by the observer on the platform is

$f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz$

Therefore the overall shift in frequency is

$\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz$

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

$v=\lambda f$

where

v is the speed of the wave

$\lambda$ is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

$\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m$

5 0

3 years ago

A parallel beam of light of wavelength 4.5 x 10^-7 m is incident on a pair of slits that are 5.0 x 10^-4 m apart. The interferen

RSB [31]

Answer:

1.8x10⁻³m

Explanation:

From the question above, the following information was used to solve the problem.

wavelength λ = 4.5x10⁻⁷m

Length L = 2.0 meters

distance d = 5 x 10₋⁴m

ΔY = λL/d

= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m

= 0.00000045 / 0.0005

= 0.0000009/0.0005

= 0.0018

= 1.8x10⁻³m

from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m

thank you!

3 0

3 years ago

Tangential velocity 2. Parabolic pathway 3. Projectile 4. Centripetal acceleration 5. Centripetal force a. acceleration towards

Stels [109]

1. Tangential velocity:

e) the instantaneous velocity of a body moving in a circular path.

2. Parabolic pathway

c. a curved path followed by projectiles

3. Projectile

d) an object projected through space, traveling in two dimensions, that accelerates vertically due to gravity.

4. Centripetal acceleration

a) acceleration towards the center caused by the centripetal force

5. Centripetal force

b) a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the center

4 0

3 years ago