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3 years ago

5

You have a frustrating day and just feel like screaming; you grab a pillow to scream into. Which interaction of sound would take

place here (Refraction, reflection, and absorption) ,and how did the sound waves move?

1 answer:

klemol [59]3 years ago

3 0

Absorption Because the sound waves go into the pillow. Not refecting or going through

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9. How much energy does a 100 W light bulb use in half an hour? If the light bulb converts

yulyashka [42]

The energy used by the light bulb in half an hour is 180000 J and the amount of thermal energy generated is 158400 J.

What is Energy?

Energy is the ability or the capacity to do work.

To calculate the energy of the light bulb we use the formula below

Formula:

E = Pt.......... Equation 1

Where:

E = Energy used by the bulb in a half-hour
P = Power of the bulb
t = Time

Given:

P = 100 W
t = 1/2 hour = 30 minutes = (30×60) = 1800 seconds

Substitute these values into equation 1

E = (100×1800)
E = 180000 J

If the light converts 12% of electric energy to light energy, then 88% of the energy is used to generate thermal energy

Therefore,

Thermal energy = (180000×88/100) = 158400 J

Hence, the energy used by the light bulb in half an hour is 180000 J and the amount of thermal energy generated is 158400 J.

Learn more about energy here: brainly.com/question/21927255

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6 0

2 years ago

A piece of plastic is uniformly charged with surface charge density eta1. The plastic is then broken into a large piece with sur

Amiraneli [1.4K]

C is the answer for this problem

3 0

3 years ago

Alexis is a scientist who is studying solid-state physics. Which activity would she most likely do as a part of this research? O

Ann [662]

Answer:

pretty sure its studying the atomic structure of a solid carbon dioxide. so c

Explanation:

8 0

3 years ago

A race car has a centriple acceleration of 15.625 m/s as it goes around a curve if the curve is a circle with radius 40 m what’s

Elanso [62]

Answer:

25 m/s

Explanation:

Centripetal acceleration is the square of the tangential velocity divided by the radius.

a = v² / r

15.625 m/s² = v² / (40 m)

v² = 625 m²/s²

v = 25 m/s

The speed of the car is 25 m/s.

7 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$

5,061.96 J = 21.5 kg × v₂² + 53. $\overline 3$ kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0

3 years ago