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3 years ago

What would seasons on the earth be like if the axis weren't tilted?

Chemistry

2 answers:

Morgarella [4.7K]3 years ago

7 0

If the Earth wasn't tilted on the axis, there would be NO seasons and humanity would be suffering.

Hope this helps,

Davinia.

Taya2010 [7]3 years ago

5 0

ANSWER:

oww this is an easy question, if the earth wasnt tilted on its axis we would have NO season!

~batmans wife dun dun dun...aka ~serenitybella

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A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

4 years ago

How much energy (heat) is required to convert 52.0 g of ice at –10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C Spe

dexar [7]

Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ

Explanation:

Using this formular, q = [mCpΔT] and = [nΔHfusion]

The energy that is needed in the different physical changes is thus:

The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:

q = [mCpΔT]

q = 52.0 x 2.09 x 10

q = 1.09 kJ

While from 0°C to 100°C is calculated as:

q = [mCpΔT]

q = 52.0 x 4.18 x 100

q = 21.74 kJ

And for fusion at 0°C is called Heat of fusion and would be given as:

q = n ΔHfusion

q = 52.0 / 18.02 x 6.02

q = 17.38 kJ

And that required for vaporization at 100°C is called Heat of vaporization and it's given as:

q = n ΔHvaporization

q = 52.0 / 18.02 x 40.7

q = 117.45 kJ

Add up all the energy gives 157.8 kJ

5 0

3 years ago

The mass of an electron is about 9.11 × 10 -9 kg., write this whole number

Pepsi [2]

Answer:0.00000009

Explanation:

9.11 × （1/1000000000）=0.00000009

8 0

3 years ago

A solution is prepared by dissolving 2.61 g ethanol (C2H5OH) in 16.2 g water. What is the molality (mol/kg) of C2H5OH in the sol

stiks02 [169]

Answer:

3.50 molal

Explanation:

Molality → Moles of solute / kg of solvent.

Let's convert the solvent's mass from g to kg

16.2 g . 1kg / 1000 g = 0.0162 kg

Let's determine the moles from the solute

2.61 g . 1 mol / 46 g = 0.0567 moles

Molality → 0.0567 mol / 0.0162 kg = 3.50 m

7 0

4 years ago

What is the weight on earth of a girl with a mass of 26 kg? N

umka2103 [35]

255 Newtons hope this helps

8 0

3 years ago