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3 years ago

Answer please, anyone?

Physics

1 answer:

liq [111]3 years ago

7 0

Physical education is one of if not the most important aspect of school, and something we can immediately transfer over to our lives. Personally, I think physical education is a must in every school and every grade level. Children are becoming more obese due to the delicious taste of fast food and how easy it is to sit on video games for hours. Kids need to know that you should not be eating at mcdonald’s every day, and instead, meal prepping. The benefits of nutrition are phenomenal including but not limited to, increased muscle size and mass, a decrease chance of heart attack, decreased chance of major artery failure etc. If you take care of your body now, you won’t have to face the consequences when you are older, and if you adopt a healthy, active lifestyle now, not only will your heart thank you, but so will your future self, for those extra few years of life you got to spend happy, and fulfilled. — I hope this is what you were looking for!

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The normal boiling point of a certain liquid is , but when of urea () are dissolved in of the solution boils at instead. Use thi

Kruka [31]

Answer:

100 Degrees is boiling point.

Explanation:

5 0

3 years ago

The part of the atom that carries no electric charge

AlexFokin [52]

The answer is: A Neutron

7 0

3 years ago

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago

An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar

igomit [66]

An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...

7 0

3 years ago

The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont

Elza [17]

Answer:

1923 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

3 0

3 years ago