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podryga [215]
3 years ago
10

All of the following are examples of mechanical energy except:

Physics
1 answer:
Paladinen [302]3 years ago
8 0
The answer is B. The nuclei of elements.
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a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
3 years ago
You place a piece of aluminum at 250.0∘C ∘ C in 9.00 kg k g of liquid water at 20.0∘C ∘ C . None of the water boils, and the fin
Mice21 [21]

Answer:

m₁ = 0.37 kg

Explanation:

According to Law of conservation of energy:

Heat Lost by Aluminum = Heat Gained by Water

m₁C₁ΔT₁ = m₂C₂ΔT₂

where,

m₁ = mass of piece of aluminum = ?

C₁ = specific heat capacity of aluminum = 900 J/kg.°C

ΔT₁ = Change in temperature of aluminum = 250°C - 22°C = 228°C

m₂ = mass of water = 9 kg

C₂ = specific heat capacity of water = 4200 J/kg.°C

ΔT₁ = Change in temperature of aluminum = 22°C - 20°C = 2°C

Therefore,

m₁(900 J/kg.°C)(228 °C) = (9 kg)(4200 J/kg.°C)(2°C)

m₁ = (75600 J)/(205200 J/kg)

<u>m₁ = 0.37 kg</u>

5 0
3 years ago
Compared to a soft sound, a loud sound at the same pitch....(check all true statements)
Sladkaya [172]

Answer:

One might think of a plucked guitar string - the sound would depend on the original amplitude of the disturbance -

Speed and velocity would still be the same

(b) is correct because the energy transfer depends on the original energy applied.

3 0
2 years ago
The ____ of emitted light depends directly on the energy between the two orbits and this determines the color of the light
Anna007 [38]

The <u>frequency</u> of emitted light is directly proportional to the energy between the two orbits and this determines the color of the light.

<h3>What is light?</h3>

Light can be defined as a form of electromagnetic waves that does not require any medium for its propagation. This ultimately implies that, light is a form of wave that is generally referred to as a visual signal because it can be seen with the eyes.

According to the model of light wave, the <u>frequency</u> of emitted light is directly proportional to the energy between the two orbits and this determines the color of the light.

<em>In conlcusion, the </em><u><em>frequency</em></u><em> of emitted </em><em>light</em><em> is highly dependent on the </em><em>energy</em><em> between the two (2) </em><em>orbits</em><em>.</em>

Read more on energy here: brainly.com/question/1242059

7 0
3 years ago
3. A rocket is launched at an angle of 53 degrees above the 1 point
irina1246 [14]

Answer:

24,000 m

Explanation:

First find the rocket's final position and velocity during the first phase in the y direction.

Given:

v₀ = 75 sin 53° m/s

t = 25 s

a = 25 sin 53° m/s²

Find: Δy and v

Δy = v₀ t + ½ at²

Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²

Δy = 7736.8 m

v = at + v₀

v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)

v = 559.0 m/s

Next, find the final position of the rocket during the second phase (as a projectile).

Given:

v₀ = 559.0 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 15945.5 m

The total displacement is:

7736.8 m + 15945.5 m

23682.2 m

Rounded to two significant figures, the maximum altitude reached is 24,000 m.

3 0
3 years ago
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