The table below shows the level of carbon dioxide in the atmosphere for a period of 50 years.


What conclusion can you make about how the carbon dioxide level has changed over time?

The level has not changed.

The level of CO2 has declined.

The level of CO2 has risen.The table below shows the level of carbon dioxide in the atmosphere for a period of 50 years.

Answer:

536.56 m/s

Explanation:

We'll begin by calculating the momentum of the Porsche. This can be obtained as follow:

Mass (m) of Porsche = 1361 kg

Velocity (v) of Porsche = 26.82 m/s

Momentum of Porsche =?

Momentum = mass × velocity

Momentum = 1361 × 26.82

Momentum of Porsche = 36502.02 Kgm/s

Finally, we shall determine the velocity you need to be running with in order to have the same momentum as the Porsche. This can be obtained as follow:

Your Mass = 68.03 kg

Your Momentum = Momentum of Porsche = 36502.02 Kgm/s

Your velocity =?

Momentum = mass × velocity

36502.02 = 68.03 × velocity

Divide both side by 68.03

Velocity = 36502.02 / 68.03

Velocity = 536.56 m/s

Thus you must be running with a speed of 536.56 m/s in order to have the same momentum as Porsche.

Answer:

The mass of the Planet X is 9.595 x 10²⁴ kg.

Explanation:

mass of the object, m = 0.5 kg

radius of the Planet X, r = 4 x 10⁶ m

weight of the object, W = F = 20 N

let the mass of the Planet  X = mₓ

Apply Newton's gravitational law;

[tex]F = \frac{Gmm_x}{r^2} \\\\m_x = \frac{Fr^2}{Gm} \\\\m_x = \frac{(20)(4\times 10^6)^2}{6.67 \times 10^{-11} \ \times \ 0.5} \\\\m_x = 9.595 \times 10^{24} \ kg[/tex]

Therefore, the mass of the Planet X is 9.595 x 10²⁴ kg.

Answer:

Distance and time.

Explanation:

Speed=Distance/time

The two factors which we need in order to calculate the speed of an object are the distance covered by the object and the time taken to cover that distance.

Speed is the rate of change of position of an object in any direction. Speed is a scalar quantity as it has only magnitude and no direction. It is measured as the ratio of the distance covered by an object to the time taken in which the distance was covered by that object.

Speed has the dimension of distance covered by the time taken. Thus, the SI unit of speed is the combination of the basic units of distance and the basic unit of Time. Thus, the SI unit of speed is meter per second (m/s).

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Answer:

zero

Explanation:

no distance has moved while holding the sheet...so no distance means no workdone..no workdone means no power...

Answer:

c. joules per gram degree Celsius

Explanation:

edg 2021

Answer:Liquid water is denser than Ice.

Ice float on top of liquid water.

Explanation:

Just trust me

Answer:

61.9°

Explanation:

The formula for calculating the workdone is expressed as

Workdone = Fdsin theta

F is the force applied on the crate

d is the distance covered

theta is the angle that the rope makes with the horizontal

Given

F = 40N

d = 7m

Workdone = 247J

Substituting into the formula:

247 = 40(7)sin theta

247 = 280sin theta

sin theta = 247/280

sin theta = 0.88214

theta = arcsin(0.88214)

theta = 61.9°

Hence the angle that the rope makes with the horizontal is 61.9°

Answer:

0J

Explanation:

No work is being done on the wall by the man pushing on it.

 Given parameters:

Mass of man  = 75kg

Mass of wall  = 500000kg

Time  = 250s

Unknown:

Work done  = ?

Solution:

Work done is the force applied on a body that moves it along a particular path.

For work to be done, distance must be move or displacement must occur.

Since the wall is not moving the distance is 0;

    Work done  = Force x distance

Since distance is 0m, work done is 0J

The work done on the wall by the man is 0 J.

To calculate the amount of work done by the man, we use the formula below.

Formula:

Where:

from the question,

Given:

Substitute these values into equation 1

Hence, the work done on the wall by the man is 0 J.

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Answer:

a) The diagrams can be seen in the picture attached

(b) By comparing the diagrams we can conclude that the resultant R₁ = R₂ = R₃

Further explanation

Vector is quantity that has magnitude and direction.

One example of a vector is acceleration.

Acceleration is rate of change of velocity.

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

This problem is about Vector and Vector Diagram.

Given:

Vector A = -200 j

Vector B = -250 i

Vector C = (150 sin 30.0°) i + (150 cos 30.0°) j = 75 i + 75√3 j

Unknown:

R₁ = A + B + C = ?

R₂ = B + C + A = ?

R₃ = C + B + A = ?

Solution:

R₁ = A + B + C  = (-200 j) + (-250 i) + (75 i + 75√3 j)

R₁ = -175i + (75√3 - 200)j

R₂ = B + C + A = (-250 i) + (75 i + 75√3 j) + (-200 j)

R₂ = -175i + (75√3 - 200)j

R₃ = C + B + A  = (75 i + 75√3 j) + (-250 i) + (-200 j)

R₃ = -175i + (75√3 - 200)j

From the results above, it can be concluded that the resultants above produce the same results. This can be confirmed from the diagrams in the attachment.

Explanation:

Answer:

If the heavy load had been lifted more slowly, the work done on the load would have been the same.

Explanation:

Work done on an object is given as;

W = Fd

where;

F is the force applied on the object

d is the displacement of the object

for the given question, the applied force on the load = mg (mass of the load multiplied by acceleration due to gravity).

Also, the displacement of the object = vertical height the load was lifted.

W = mgh

The work done on the load is independent of time.

Thus, if the heavy load had been lifted more slowly, the work done on the load would have been the same.

A person lifting a heavy load to a vertical height of 2.0 m in 3 seconds does the same work as if he/she lifts it in 6 s.

A person lifts a heavy load to a vertical height of 2.0 m in 3 seconds.

We want to compare the work done with the one that he/she would have done if the process had taken 6 seconds.

In physics, work (W) is the energy transferred to or from an object via the application of force (F) along a displacement (s).

W = F × s

Given the displacement is the same (2.0 m) and the force needed is also the same (weight of the object), the work is the same for both processes.

A person lifting a heavy load to a vertical height of 2.0 m in 3 seconds does the same work as if he/she lifts it in 6 s.

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Answer: 432m

Explanation:

Convert 3 min to seconds

1 min = 60 sec

3 min = 180 sec

Multiply the speed times time to get distance.

2.4 x 180 = 432m

Answer:

4770 N

Explanation:

Momentum is the product of mass and velocity and force is the change in momentum divided by change in time.

Given from the question;

Mass of arrow= 0.53 kg

Velocity of arrow = 63 m/s

Initial velocity of arrow = 0 m/s

Change in time = 0.007 s

Finding momentum after the arrow is released as;

p=m*v

p= 0.53 * 63

p= 33.39 kg*m/s

Force is the change in momentum divided by change in time;

F= 33.39 / 0.007

F= 4770 N

Answer:

s = 2.79 m

Explanation:

Given that,

A ball have a speed of 7.4 m/s just before it hits the ground.

Initial velocity of the ball was 0 (at rest)

We need to find the height from where the ball was dropped. It means we need to find the distance covered by it. Let it is h.

Using the third equation of motion to find it as follows :

[tex]v^2-u^2=2as\\\\s=\dfrac{v^2}{2g}\\\\s=\dfrac{(7.4)^2}{2\times 9.8}\\\\s=2.79\ m[/tex]

So, the ball is dropped from a height of 2.79 m.

The height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground is 2.79m

According to the third equation of motion;

[tex]v^2=u^2+2as[/tex]

where;

v is the final velocity = 7.4m/s

Initial velocity = 0m/s

s is the distance

a = g= 9.8m/s²

Substitute the values into the formula to have:
[tex]7.4^2=0^2+2(9.8)s\\54.76 + 19.6s\\s=\frac{54.76}{19.6}\\ s=2.79m[/tex]

Hence the height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground is 2.79m

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Answer:

y = 14238 m.    the height of the rocket is much less than this distance therefore the plan will not work.

Explanation:

Let's analyze this exercise, so that the Martian's plan works, the vertical height of the body must be zero when it is more than half of the way to the planet Mars, this is so that Mars attracts it and can arrive.

Let's calculate the maximum height of the launch

          [tex]v_{y} ^2 = v_{oy}^2 - 2 g y[/tex]

at the highest point [tex]v_{y}[/tex] = 0

          y = v_{oy}² / 2g

          y = (v₀ sin θ)² / 2g

let's calculate

          y = (1250 sin 25)² /2 9.8

          y = 14238 m

In the exercise, indicate that the distance to Mars is h = 321770 m, half of this distance is

          h / 2 = 160885 m

therefore the height of the rocket is much less than this distance therefore the plan will not work.

The height reached is low, so it is not necessary to take into account the variation of g with height

Answer: 85J

Explanation:

From the question, we are informed that when spring was wound, 100J of work was done but 15J escaped to the surrounding as heat.

Therefore, the change in internal energy of the spring will be calculated as:

ΔU = q + w

where, q = -15J

w = 100J

ΔU = -15J + 100J

= 85J

Answer: = 2,496 N

Explanation:

Force is the push or pull on an object and it is calculated as:

= Mass * Acceleration

= 1,248 * 2.0

= 2,496 N

Answer:

3743.489 kg

Explanation:

F_g = 591 N

G = 6.674x10^-11 constant of gravity

m_1 = 95 kg

m_2 = unknown

r = 4990*1000 =

F_g = G[(m_1*m_2)/r^2]

591 N = 6.674x10^-11[(95*m_2)/4990^2]

8.855 = [(95*m_2)/4990^2]

355631.472 = 95*m_2

m_2 = 3743.489 kg

The  mass of the planet, which has a radius of 4990 km, is  1.81×10²³kg.

Newton's law of universal gravitation states that

The force of attraction between any two bodies is inversely proportional to the square of the distance between them and directly proportional to the product of their masses.

Given parameters:

Mass of the object: m = 95 kg

Radius of the planet: r = 4990 km = 4990 × 1000 m.

Weight f the object: F= 591 N

We know that: universal gravitational constant: G = 6.674x10^-11 SI unit.

We have to find: mass of the planet: M = ?

Now, F= GMm/r^2

591 N = 6.674x10^-11[(95×M)/(4990×1000)^2]

⇒ M =1.81×10²³kg

Hence, mass of the planet is  1.81×10²³kg.

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Answer:

[tex]D= -0.42km[/tex]

Explanation:

From the question we are told that

Drifting right with speed 5.0m/s

The motor runs for 6.0 seconds

Leftward acceleration of magnitude 4.0 m/s squared

Generally the equation [tex]V=ut+1/2at^2[/tex] can be used here

[tex]V=ut+1/2at^2[/tex]

Mathematically solving with the newton equation above we have that

   [tex]D=5*6 + \frac{1}{2} (-4)*6^2[/tex]

   [tex]D=30-72[/tex]

   [tex]D=-42m[/tex] [tex]or -0.42km[/tex]

Therefore having this the Displacement is [tex]D= -0.42km[/tex] leftward

Answer:

one could freeze and the second would thaw

Explanation:

sorry if its wrong

The two examples of how land can have a dramatic change in temperature

is during freezing and thawing.

Cold temperatures which is common during the winter period is

characterized by the formation of snow and freezing of smaller water

bodies.

There may be a short phase in which there is relative sunlight which melts

the frozen substances thereby forming liquids . This is usually as result of

the temperature being on the increase in the atmosphere.

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Answer:

3.44× 10⁻¹⁹Joules

Explanation:

Energy of the wavelength is expressed using the formula:

E = hc/λ

h is the Planck constant

c is the velocity of light

λ is the wavelength

Given

h = 6.63 × 10^-34 m² kg / s

c = 3×10⁸ m/s

λ = 579nm = 579 × 10⁻⁹m

λ = 5.79× 10⁻⁷m

Substitute the given values into the formula

E = hc/λ

E = (6.63 × 10⁻³⁴× 3×10⁸)/5.79× 10⁻⁷

E = 19.89× 10⁻³⁴⁺⁸/5.79× 10⁻⁷

E = 19.89× 10⁻²⁶/5.79× 10⁻⁷

E = 3.44× 10⁻²⁶⁺⁷

E = 3.44× 10⁻¹⁹Joules

Hence the energy of this wavelength of light is 3.44× 10⁻¹⁹Joules

Answer:

41.02m

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity  = 96km/hr

Time taken  = 3.07s

Unknown:

Distance traveled by the time the final speed was achieved = ?

Solution:

To solve this problem, we first find the acceleration of the car;

     Acceleration  = [tex]\frac{v - u }{t}[/tex]

v is the final velocity

u is the initial velocity

t is the time taken

  Now convert the the final velocity to m/s;

          96km/hr to m/s;

               1 km/hr  = 0.278m/s

               96km/hr  = 96 x 0.278 = 26.7m/s

Now;

 Acceleration  = [tex]\frac{26.7 - 0}{3.07}[/tex]   = 8.69m/s²

So;

   v²  = u²  + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

             26.7²  = 0²  + 2 x 8.69 x s

             712.89  = 17.38s

                  s  = 41.02m