Based on what you know, please list at least 4 characteristics of weather. IDK IF RIGHT SUBJECT

PLSSSSSSSS SOMEONE HELP ME WITH THIS ONE!!

zloy xaker [14]

Answer:

D

Explanation:

At the other end of the spectrum toward red, the wavelengths are longer and have lower energy.

8 0

3 years ago

Write a sentence about sky teaches

Marina CMI [18]

Sky teaches us many things which cannot be taught by anyone else.

4 0

4 years ago

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For a particular reaction, Δ H ∘ = − 93.8 kJ and Δ S ∘ = − 156.1 J/K. Assuming these values change very little with temperature,

svetoff [14.1K]

To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity

At constant temperature and pressure, the change in Gibbs free energy is defined as

$\Delta G = \Delta H - T\Delta S$

Where,

H = Entalpy

T = Temperature

S = Entropy

When the temperature is less than that number it is negative meaning it is a spontaneous reaction. $\Delta G$ is also always 0 when using single element reactions. In numerical that implies $\Delta G = 0$

At the equation then,

$\Delta G = \Delta H - T\Delta S$

$0 = \Delta H - T\Delta S$

$\Delta H = T\Delta S$

$T = \frac{\Delta H}{\Delta S}$

$T = \frac{-93.8kJ}{-156.1J/K}$

$T = \frac{-93.8*10^3J}{-156.1J/K}$

$T = 600.89K$ }

Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K

5 0

3 years ago

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0

3 years ago

A young owl forgets how to fly for a moment and is free falling towards the ground. Before crashing, he remembers how to fly and

Oksi-84 [34.3K]

did you figure it out

5 0

4 years ago

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