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3 years ago

The use of a beat or a rhythm to remember something is

2 answers:

5 0

The answer is Music Mnemonics. Mnemonic is originally coming from the Ancient Greek word means “memory”. It is the study and development of systems for improving and assisting the memory. Mnemonic techniques are more specific memory aids. Music Mnemonic is a way we used to remember song lyrics and rhythmic movement.

Aleonysh [2.5K]3 years ago

5 0

Answer: C music mnemonics

Explanation: Its on apex

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Winds that blow from the north and south poles

Gennadij [26K]

Winds that blow from the north and south poles would be called katabatic winds. I'm not sure if I spelled that right, but that's the answer I hope.

8 0

2 years ago

The major contributions of Maury included:

scoray [572]

The best and most correct answer among the choices provided by your question is the second choice.

The major contributions of Maury included mapping the ocean bottom.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

6 0

2 years ago

Which branch of physics deals with the study of force, energy, and motion?

yawa3891 [41]

The branch of physics that deals with the study of force energy and motion is classic mechanics

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2 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

a vector a has components a x equals -5.00 m in a y equals 9.00 meters find the magnitude and the direction of the vector

Murrr4er [49]

Answer:

The magnitude = 10.30 m

The direction of the vector proceeds at angle of 119.05°

Explanation:

Given that:

A vector $\bar A$ has component $A_x$ = -5 m and $A_y$ = 9 m

The magnitude of vector $\bar A$ can be represented as:

$\bar A$ = $\sqrt{A_x^2 + A_y^2}$

$\bar A$ = $\sqrt{(-5)^2 + (9)^2}$

$\bar A$ = $\sqrt{25 + 81}$

$\bar A$ = $\sqrt{106}$

$\bar A$ = 10.30 m

If we make $\bar A$ an angle $\theta$ with y- axis:

Then; tan $\theta$ = $\frac{A_x}{A_y}$

tan $\theta$ = $\frac{5}{9}$

tan $\theta$ = 0.555

$\theta$ = tan⁻¹ (0.555)

$\theta$ = 29.05°

Angle with positive x-axis = 90 + $\theta$

= 90° + 29.05°

= 119.05°

5 0

3 years ago