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vazorg [7]
3 years ago
9

What is a major contributor of greenhouse gases in the atmosphere?

Physics
1 answer:
ASHA 777 [7]3 years ago
6 0

Answer:

a. burning of fossil fuel

Explanation:

Greenhouse effect is the trapping of the sun infrared rays in the outermost layer of the earths atmosphere due to the accumulation of some harmful gasses. This gases depletes the ozone layer

The major contributor of greenhouse gases is the burning of fossil fuels. Carbon dioxides are released into the atmosphere and leads to global warming and climatic changes per time

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A racecar is equipped with a computer that records the reading on its speedometer every second during a race. If you graph this
Tanya [424]
Because it records speed of the car at a certain time, the independent variable should be time and dependent would be speed or velocity. Since it's taken every second, it would be considered instantaneous velocity, which is D.
4 0
3 years ago
Why do people hate on people and hate country people. are you country
Dmitry_Shevchenko [17]

Answer:

why do people hate on people and hate country people. are you country

because they only think about. themselves and they also think that they are always right .

it's my point of view

hope it is helpful to you

3 0
3 years ago
Read 2 more answers
A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.
N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2

(c) Moment of inertia :

The net torque acting on it is, \tau=I\alpha, I is the moment of inertia

Also, \tau=Fr

So,

I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2

Hence, this is the required solution.

3 0
3 years ago
Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const
bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

S = ut + \frac{1}{2}at^{2}

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

#SPJ1

7 0
2 years ago
A vat of nitrogen at its boiling point (-196 degrees C) absorbs 384000 J of heat. How much mass of nitrogen burns off? (Unit = k
masya89 [10]

Answer:

1.92 kg of nitrogen.

Explanation:

The following data were obtained from the question:

Heat absorbed (Q) = 384000 J

Note: Heat of vaporisation (ΔHv) of nitrogen = 5600 J/mol

Next, we shall determine the number of mole of nitrogen that absorbed 384000 J.

This is illustrated below:

Q = mol·ΔHv

384000 = mole of N2 x 5600

Divide both side by 5600

Mole of N2 = 384000/5600

Mole of N2 = 68.57 moles

Next, we shall convert 68.57 moles of nitrogen, N2 to grams.

This can be obtained as follow:

Molar mass of N2 = 2 x 14 = 28 g/mol.

Mole of N2 = 68.57 moles.

Mass of N2 =..?

Mole = mass /molar mass

68.57 = mass of N2 /28

Cross multiply

Mass of N2 = 68.57 x 28

Mass of N2 = 1919.96 g

Finally, we shall convert 1919.96 g to kilograms.

This can be achieved as shown below:

1000g = 1 kg

Therefore,

1919.96 g = 1919.96/1000 = 1.92 kg.

Therefore, 1.92 kg of nitrogen were burned off.

3 0
3 years ago
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