2.3*10^14 and 5.4*10^-3

Which body of water will have a greater influence on an area?

Andrei [34K]

Answer:

The correct option is A

Explanation:

Water from a river is used for many activities in a community. These activities could include (but not limited to) tourism, drinking for animals, local transport, irrigation for nearby farming, recreation (as in swimming), habitat for some living organisms among others. Rivers are not limited by what limits the influence of oceans such as taste (it's saltiness, which cannot be used in farming also) and wave current.

5 0

3 years ago

How many electrons in each orbit?

sveticcg [70]

In 1st orbit 2
2nd 8
3rd 10
f orbital has 16

4 0

3 years ago

Once a slide is prepared and placed onto the microscope, the magnification and focus need to be altered. Describe how to change

Harlamova29_29 [7]

Answer:

我實際上不知道答案，我只是為了點數而這樣做，哈哈，祝你好運哈哈

Explanation:

我實際上不知道答案，我只是為了點我實際上不知道答案，我只是為了點數而這樣做，哈哈，祝你好運哈哈數而這樣做，哈哈，祝你好運哈哈

5 0

3 years ago

Read 2 more answers

A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the fl

diamong [38]

Answer:

$\large \boxed{\text{98 m$\cdot$s}^{-2}}$

Explanation:

The formula for the velocity of the ball is

$v = \sqrt{2gh}$

1. Velocity at time of impact

$v = -\sqrt{2 \times 9.807 \times 1.20} = -\sqrt{23.54} = -\textbf{4.85 m/s}$

2. Velocity on rebound

The ball has enough upward velocity to reach a height of 0.86 m.

$v = \sqrt{2 \times 9.807 \times 0.86} = \sqrt{16.87} =\textbf{4.11 m/s}$

3. Acceleration

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{4.11 - (-4.85)}{ 0.091} = \dfrac{8.96 }{0.091} =\textbf{98 m$\cdot$s}^{\mathbf{-2}}\\\\\text{The acceleration while the ball is in contact with the floor is $\large \boxed{\textbf{98 m$\cdot$s}^{\mathbf{-2}}}$}$

5 0

3 years ago

Based on the three formulas shown, use one of them to solve for the purple yellow and red box and explain how you did it.

zysi [14]

P = 11.133 atm (purple)

T = -236.733 °C(yellow)

n = 0.174 mol(red)

<h3>Further explanation </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:

Boyle's law at constant T, P = 1 / V
Charles's law, at constant P, V = T
Avogadro's law, at constant P and T, V = n

So that the three laws can be combined into a single gas equation, the ideal gas equation

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

Purple box

T= 10 +273.15 = 373.15 K

V=5.5 L

n=2 mol

$\tt P=\dfrac{nRT}{V}\\\\P=\dfrac{2\times 0.08205\times 373.15}{5.5}\\\\P=11.133~atm$

Yellow box

V=8.3 L

P=1.8 atm

n=5 mol

$\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{1.8\times 8.3}{5\times 0.08205}\\\\T=36.42~K=-236.733^oC$

Red box

T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

$\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol$

3 0

3 years ago