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3 years ago

Why does the Law of Conservation of Matter/Mass/Energy support climate change research?

1 answer:

4 0

The law of conservation of energy supports climate change research because the energy we use is released into the atmosphere, creating loop of continual absorption and releasing of heat.

<h3>Law of conservation of energy</h3>

The law states that energy can neither be created nor destroyed but can always be converted from one form to another.

In light of this, any energy used by humans or natural processes is not destroyed. Instead, they are converted into another form or released into the atmosphere.

In the atmosphere, absorption and release of heat energy by greenhouse gases and irradiation create a loop. This loop causes a net heating or cooling of the planet, a phenomenon that affects the climate.

More on the law of conservation of energy can be found here: brainly.com/question/20971995

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The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21

frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is $h(t) = 3+21.5\cdot t-4.9\cdot t^{2}$ after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 seconds and after 4 seconds, (b) When does the projectile reach its maximum height? (c) What is the maximum height? (d) When does it hit the ground? (e) With what velocity does it hits the ground?

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

$v(t) = 21.5-9.8\cdot t$ (Eq. 1)

Where $v(t)$ is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

$v(2) = 21.5-9.8\cdot (2)$

$v(2) = 1.9\,\frac{m}{s}$

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

$v(4) = 21.5-9.8\cdot (4)$

$v(4) = -17.7\,\frac{m}{s}$

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for $t$ :

$21.5-9.81\cdot t = 0$

$t = 2.192\,s$

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

$h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}$

$h (2.192) = 26.584\,m$

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for $t$ . That is:

$3+21.5\cdot t-4.9\cdot t^{2} = 0$

Roots are found by means of Quadratic Formula:

$t_{1}\approx 4.523\,s$ and $t_{2}\approx -0.135\,s$

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

$v(4.523) = 21.5-9.81\cdot (4.523)$

$v(4.523) = -22.871\,\frac{m}{s}$

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

5 0

4 years ago

When you multiply (2.2500x10^3) x (3.00 x10^5) =675,000,000 report this answer to decimal using scientific notation

zavuch27 [327]

Answer:

$6.75\times 10^{8}$

Explanation:

(2.2500x10^3) x (3.00 x10^5)

First add the power values since all are to power 10 hence this will be 3+5=8

Then multiply the digit intergers ie 2.25*3=6.75

Therefore, in scientific notation this will be represented as $6.75\times 10^{8}$

8 0

4 years ago

Does the size of the magnetic field change with the size of the magnet?

mestny [16]

Answer:

Not all of the time, it can also depend on the strength of the magnet.

Explanation:

say you have a small very strong magnet, that might work just as well as a large but week magnet.

4 0

3 years ago

Suppose there are two identical gas cylinders. One contains the monatomic gas krypton (Kr), and the other contains an equal mass

tensa zangetsu [6.8K]

Answer:

Explanation:

Let equal mass of Ne and Kr be m gm

no of moles of Ne and Kr will be m / 20 and m / 84 ( atomic weight of Ne and Kr is 20 and 84 )

Let the pressure and volume of both the gases be P and V respectively .

The temperature of Ne be T₁ and temperature of Kr be T₂.

For Ne

PV = (m / 20) x R T₁

For Kr

PV = (m / 84) x R T₂

T₁ / T₂ = 84 / 20

We know that

average KE of an atom of mono atomic gas = 3 / 2 x k T

k is boltzmann constant and T is temperature .

KEKr/KENe = T₂ / T₁

= 20 / 84

4 0

3 years ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

4 years ago