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3 years ago

Under what atmospheric condition is fog most often formed in the san joaquin valley?

1 answer:

4 0

It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway,here is the answer. The atmospheric condition in which <span>fog is most often formed in the san Joaquin valley is stable stability. Hope this helps.</span>

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A sample of helium (He) occupies 8.0 liters at 1 atm and 20.0◦C. What pressure is necessary to change the volume to 1.0 liters a

nevsk [136]

Apply the combined gas law

PV/T = const.

P = pressure, V = volume, T = temperature, PV/T must stay constant.

Initial PVT values:

P = 1atm, V = 8.0L, T = 20.0°C = 293.15K

Final PVT values:

P = ?, V = 1.0L, T = 10.0°C = 283.15K

Set the PV/T expression for the initial and final PVT values equal to each other and solve for the final P:

1(8.0)/293.15 = P(1.0)/283.15

P = 7.7atm

7 0

3 years ago

Describe what it means if a chemical or physical property is periodic

vesna_86 [32]

It means that you consider the elements as a list organized by atomic number, the property is seen to repeat over and over as you move through that list.

6 0

3 years ago

Why in example multiplying the length by 2

dolphi86 [110]

What??????? idk!!!!!

3 0

2 years ago

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stoc

vivado [14]

Answer:

Δ $R_{e}$ = 84 Ω, $R_{e}$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_{e}$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_{e}$ = 1 / R₁ + 1 / R₂

1 / $R_{e}$ = 1/500 + 1/2000 = 0.0025

$R_{e}$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_{e}$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_{e}$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_{e}$ / $R_{e}$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_{e}$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_{e}$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_{e}$ = 0.21 400

Δ $R_{e}$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_{e}$ = (40 ± 8) 10¹ Ω

6 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

2 years ago