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White raven [17]
3 years ago
5

An interglacial is a period of time that occurs between Ice Ages. It’s marked by warmer temperatures and milder climates. During

previous interglacials, the CO2 levels in the atmosphere were about 280 parts per million (ppm). In 2013, they were approximately 400 ppm. Rounding to the nearest percent, what is the percent increase in CO2 between these two time periods?
Physics
1 answer:
Korolek [52]3 years ago
3 0

Answer:

The correct answer is - 43%.

Explanation: The increase in CO2 between these two suggested periods is approximately 43%. Even though it is a natural process that the CO2 levels vary in the atmosphere, still this is not the same case nowadays. Nowadays, or rather in the past few decades, apart from the natural increase of CO2 in the atmosphere, it has seen a much more increased levels because of the human activity. The industrial facilities and the vehicles, the cutting of the forests and burning the wood (there's both release of CO2 from the burning of the trees and loss of natural accumulator of the CO2), are just some of the more important human activities that contribute to a significant rise in the CO2 levels.

You might be interested in
A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

3 0
4 years ago
A student calculates the density of a copper cube to be 4.15 g/cm . If the accepted value is 8.64 g/cm the percentage error in h
Helen [10]

 The percentage error in his experimental value is -51.97%.

<h3>What is percentage error?</h3>

This is the ratio of the error to the actual measurement, expressed in percentage.

To calculate the percentage error of the student, we use the formula below.

Formula:

  • Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1

From the question,

Given:

  • Calculated value = 4.15 g/cm
  • accepted value = 8.64 g/cm

Substitute these values into equation 1

  • Error(%) = (4.15-8.64)100/8.64
  • Error(%) = -4.49(100)/8.64
  • Error(%) = -449/8.64
  • Error(%) = -51.97 %

 

Hence, The percentage error in his experimental value is -51.97%.

Learn more percentage error here: brainly.com/question/5493941

8 0
3 years ago
Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension fo
Vlad1618 [11]

Answer:

Tension T1 is less than tension T2.

T1 < T2

Explanation:

According to given data,

mass of box A ( mA) is grater than mass of box B (mB)

we can write,

m(A) > m(B)

Newton's second law states that:

Tension of object is directly proportional to the mass of the system.

T ∝ m

here Boxes A and B are being pulled to the right on a frictionless surface,

so Tension T1 generates due to the mass of box A m(A)

and Tension T2 arises due to mass of the system m(A) + m(B)

Thus tension T1 will be less than tension T2

T1 < T2

learn more about Tension force here:

<u>brainly.com/question/13175014</u>

<u />

#SPJ4

8 0
2 years ago
A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain
Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m

The amplitude is 0.5 m

Time period is given by

T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s

Number of oscillations is given by

N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595

The number of oscillations is 14.00595

For maximum speed

\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16

At x = 0.35 m

v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s

The speed of the block is 5.71314 m/s

4 0
3 years ago
Read 2 more answers
What kind of stored energy enables a stretched bow to shoot an arrow?
blsea [12.9K]
Sorry if im wrong but I'll say kinetic energy

Hope it helps:)
7 0
4 years ago
Read 2 more answers
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