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makkiz [27]
3 years ago
6

Natural gas is used for electricity generation, domestic purposes, transportation, and industrial manufacturing of fertilizers.

Methane is a byproduct of natural gas combustion. The following table reports the total U.S. methane emissions from natural gas systems in million metric tons of carbon equivalent (MMTCE): Year 1990 1995 2000 2005 2010 Total Methane (in MMTCE) 32.9 33.2 35.6 36.7 37.9 Data Courtesy of EPA Which statement most likely describes an impact of the trend shown in the table?
Physics
1 answer:
ankoles [38]3 years ago
4 0
1195 or either 33.2 or 2010
You might be interested in
Bacteria vary in size, but a diameter of 2.0 µm is not unusual. What are the volume (in cubic centimeters) and surface area (in
telo118 [61]

A = 4\pi r^2

A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2

A = 1.33*!0^{-5}MM^2

5 0
3 years ago
Gravitational notes of physics ​
Pachacha [2.7K]

Answer:

Every object in the universe attracts other object by a force of attraction, called gravitation, which is directly proportional to the product of masses of the objects and inversely proportional to the square of distance between them. This is called Law of Gravitation or Universal Law of Gravitation.

Let masses (M) and (m) of two objects are distance (d) apart. Let F be the attractional force between two masses.

Importance of The Universal Law of Gravitation

It binds us to the earth.

It is responsible for the motion of the moon around the earth.

It is responsible for the motion of planets around the Sun.

Gravitational force of moon causes tides in seas on earth.

Free Fall

When an object falls from any height under the influence of gravitational force only, it is known as free fall.

Acceleration Due to Gravity

When an object falls towards the earth there is a change in its acceleration due to the gravitational force of the earth. So this acceleration is called acceleration due to gravity.

The acceleration due to gravity is denoted by g.

The unit of g is same as the unit of acceleration, i.e., ms−2

Mathematical Expression for g

From the second law of motion, force is the product of mass and acceleration.

F = ma

For free fall, acceleration is replaced by acceleration due to gravity.

Therefore, force becomes:

F = mg ….(i)

But from Universal Law of Gravitation,

Factors Affecting the Value of g

As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator.

As we go at large heights, value of g decreases.

To Calculate the Value of g

Value of universal gravitational constant, G = 6.7 × 10–11 N m2/ kg2,

Mass of the earth, M = 6 × 1024 kg, and

Radius of the earth, R = 6.4 × 106 m

Putting all these values in equation (iii), we get:

Thus, the value of acceleration due to gravity of the earth, g = 9.8 m/s2.

Difference between Gravitation Constant (G) and Gravitational Acceleration (g)

S. No.

Gravitation Constant (G)

Gravitational acceleration (g)

1.

Its value is 6.67×10-11Nm2/kg2.

Its value is 9.8 m/s2.

2.

It is a scalar quantity.

It is a vactor quantity.

3.

Its value remains constant always and everywhere.

Its value varies at various places.

4.

Its unit is Nm2/kg2.

Its unit is m/s2.

Motion of Objects Under the Influence of Gravitational Force of the Earth

Let an object is falling towards earth with initial velocity u. Let its velocity, under the effect of gravitational acceleration g, changes to v after covering the height h in time t.

Then the three equations of motion can be represented as:

Velocity (v) after t seconds, v = u + ght

Height covered in t seconds, h = ut + ½gt2

Relation between v and u excluding t, v2 = u2 + 2gh

The value of g is taken as positive in case of the object is moving towards earth and taken as negative in case of the object is thrown in opposite direction of the earth.

Mass & weight

Mass (m)

The mass of a body is the quantity of matter contained in it.

Mass is a scalar quantity which has only magnitude but no direction.

Mass of a body always remains constant and does not change from place to place.

SI unit of mass is kilogram (kg).

Mass of a body can never be zero.

Weight (W)

The force with which an object is attracted towards the centre of the earth, is called the weight of the object.

Now, Force = m × a

But in case of earth, a = g

∴ F = m × g

But the force of attraction of earth on an object is called its weight (W).

∴ W = mg

As weight always acts vertically downwards, therefore, weight has both magnitude and direction and thus it is a vector quantity.

The weight of a body changes from place to place, depending on mass of object.

The SI unit of weight is Newton.

Weight of the object becomes zero if g is zero.

Weight of an Object on the Surface of Moon

Mass of an object is same on earth as well as on moon. But weight is different.

Weight of an object is given as,

Hence, weight of the object on the moon = (1/6) × its weight on the earth.

Try the following questions:

Q1. State the universal law of gravitation.

Q2. When we move from the poles to the equator, the value of g decreases. Why?

Q3. If two stones of 150 gm and 500 gm are dropped from a height, which stone will reach the surface of the earth first and why ?

Q4. Differentiate between weight and mass.

Q5. Why is the weight of an object on the moon 1/6th its weight on the earth??

7 0
3 years ago
When Jane drives to work, she always places her purse on the passenger’s seat. By the time she gets to work, her purse has falle
madam [21]

Answer:

the vibrations push the purse up and down very fast and gravity pushes the purse down onto the floor

Explanation: does that help

7 0
3 years ago
A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,
Maru [420]

Answer:

a) Coefficient of kinetic friction between block and surface = 0.12

b) Decrease in kinetic energy of the bullet = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = 0.541 J

Explanation:

Given,

Mass of bullet = 4.00 g = 0.004 kg

Initial velocity of the bullet = 400 m/s

Mass of wooden block = 0.65 kg

Initial velocity of the wooden block = 0 m/s (since it was initially at rest)

Final velocity of the bullet = 190 m/s

Distance slid through by the block after the collision = d = 72.0 cm = 0.72 m

Let the velocity of the wooden block after collision be v

According to the law of conservation of momentum,

Momentum before collision = Momentum after collision

Momentum before collision = (Momentum of bullet before collision) + (Momentum of wooden block before collision)

Momentum of bullet before collision = (0.004×400) = 1.6 kgm/s

Momentum of wooden block before collision = (0.65)(0) = 0 kgm/s

Momentum after collision = (Momentum of bullet after collision) + (Momentum of wooden block after collision)

Momentum of bullet after collision = (0.004×190) = 0.76 kgm/s

Momentum of wooden block after collision = (0.65)(v) = (0.65v) kgm/s

Momentum balance gives

1.6 + 0 = 0.76 + 0.65v

0.65v = 1.6 - 0.76 = 0.84

v = (0.84/0.65)

v = 1.29 m/s

The velocity of the wooden block after collision = 1.29 m/s

To obtain the coefficient of kinetic friction between block and surface, we will apply the work-energy theorem.

The work-energy theorem states that the work done in moving the block from one point to another is equal to the change in kinetic energy of the block between these two points.

The points to consider are the point when the block starts moving (immediately after collision) and when it stops as a result of frictional force.

Mathematically,

W = ΔK.E

W = workdone by the frictional force in stopping the wooden block (since there is no other horizontal force acting on the block)

W = -F.d (minus sign because the frictional force opposes motion)

d = Distance slid through by the block after the collision = 0.72 m

F = Frictional force = μN

where N = normal reaction of the surface on the wooden block and it is equal to the weight of the block.

N = W = mg

F = μmg

W = - μmg × d = (-μ)(0.65)(9.8) × 0.72 = (-4.59μ) J

ΔK.E = (final kinetic energy of the block) - (initial kinetic energy of the block)

Final kinetic energy of the block = 0 J (since the block comes to a rest)

(Initial kinetic energy of the block) = (1/2)(0.65)(1.29²) = 0.541 J

ΔK.E = 0 - 0.541 = - 0.541 J

W = ΔK.E

-4.59μ = -0.541

μ = (0.541/4.59)

μ = 0.12

b) The decrease in kinetic energy of the bullet

(Decrease in kinetic energy of the bullet) = (Kinetic energy of the bullet before collision) - (Kinetic energy of the bullet after collision)

Kinetic energy of the bullet before collision = (1/2)(0.004)(400²) = 320 J

Kinetic energy of the bullet after collision = (1/2)(0.004)(190²) = 72.2 J

Decrease in kinetic energy of the bullet = 320 - 72.2 = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = (1/2)(0.65)(1.29²) = 0.541 J

Hope this Helps!!!

4 0
3 years ago
A 5.30 m long pole is balanced vertically on its tip. What will be the speed (in meters/second) of the tip of the pole just befo
suter [353]

Answer:

v = 17.66 m/s

Explanation:

As we know that the lower end of the pole is fixed in the ground and it start rotating about that end

so here we can say that the gravitational potential energy of the pole will convert into rotational kinetic energy of the pole about its one end

so we have

mgL = \frac{1}{2}(\frac{mL^2}{3})\omega^2

so we have

\omega = \sqrt{\frac{6g}{L}}

now we have

\omega = \sqrt{\frac{6(9.81)}{5.30}}

\omega = 3.33 rad/s

now the speed of the other tip of the pole is given as

v = \omega L

v = (3.33)(5.30) = 17.66 m/s

8 0
4 years ago
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