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3 years ago

Which shows the formula for converting from degrees Celsius to degrees Fahrenheit?

Physics

2 answers:

masha68 [24]3 years ago

8 0

Explanation:

The two scales for measuring the temperature of an object are Celsius scale and Fahrenheit scale. Mathematically, it can be written as :

$^{\circ} F=(\dfrac{9}{5}\times ^{\circ}C)+32$

Where

°F = degree Fahrenheit

°C = degree Celsius

For example, if T = 30°C. It can be converted to °F as :

$^{\circ} F=(\dfrac{9}{5}\times 30)+32$

$^{\circ} F=86$

So, 30°C is equal to 86 Fahrenheit. Hence, this is the required solution.

Alina [70]3 years ago

3 0

This is a question that makes us REALLY want to know what "Which" means.

If you included a list of answer choices with your question, then the correct formula isn't on it.

The formula for converting temperature from degrees Celsius to degrees Fahrenheit is:

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vfiekz [6]

Yes that answer is correct because sometimes tides cause waves to happen

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3 years ago

Read 2 more answers

A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).

qaws [65]

To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

rms = sqrt (3 R T / M)

where

R = gas constant = 8.314 Pa m^3 / mol K

T = temperature

M = molar mass

Now we get the ratios of rms of Argon (1) to hydrogen (2):

rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))

rms1 / rms2 = sqrt (T1 M2 / T2 M1)

Since T1 = 4 T2

rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)

rms1 / rms2 = sqrt (4 M2 / M1)

and M2 = 2 while M1 = 40

rms1 / rms2 = sqrt (4 * 2 / 40)

rms1 / rms2 = 0.447

Therefore the ratio of rms is:

<span>rms_Argon / rms_Hydrogen = 0.45</span>

7 0

3 years ago

PLEASE HELP :) THANK YOU Geologic processes change Earth's surface on varying scales of space and time. They range from rapid to

Misha Larkins [42]

Answer:

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Explanation:

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3 years ago

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle

ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E = 5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0

3 years ago

If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat

Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, $m_p$ = 1.67 x 10⁻²⁷ kg

mass of electron, $m_e$ = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

$F = \frac{k(q_p)(q_e)}{r^2}$

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

$F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N$

Acceleration of proton is given by;

F = ma

$F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2$

Acceleration of the electron is given by;

$F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2$

7 0

3 years ago