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Alex777 [14]
3 years ago
15

Burning a canful of petrol or dropping it. which is likely to release most energy?​

Physics
1 answer:
Neporo4naja [7]3 years ago
5 0

Answer:

Burning of canful of petrol will release more energy

Explanation:

If you like my answer than please mark me brainliest

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PROBLEMA DI FISICA:il volume di un attizzatoio da camino in ferro, a temperatura ambiente (20°C), è 12,2 cm cubi. A contatto con
lesantik [10]

Responder:

20.3 ° C

Explicación:

<u>Según la ley de Charles</u>: <em>cuando la presión sobre una muestra de gas seco se mantiene constante, la temperatura y el volumen estarán en proporción directa. </em>

Paso uno:

datos dados

Temperatura T1 = 20 ° C

Temperatura T2 =?

Volumen V1 = 12.2 cm ^ 3

Volumen V2 = 12.4 cm ^ 3

Aplicar la relación temperatura y volumen

\frac {V_{1}}{T_{1}}=\frac {V_{2}}{T_{2}}

sustituyendo tenemos

\frac {12.2}{20}=\frac {12.4}{T_{2}}\\\\

Cruz multiplicar tenemos

T_2=\frac{20*12.4}{12.2} \\\\T_2=\frac{248}{12.2}\\\\T_2=20.3

Temperatura delle braci 20.3°C

4 0
3 years ago
Light propagate faster through medium “a” than medium “b”
dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1, n_2 are the index of refraction of the two mediums

\theta_1, \theta_2 are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

sin \theta_1 = sin \theta_2 (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

n_1\neq n_2

Therefore the only angle that can satisfy eq.(1) is

\theta_1 = \theta_2 = 0

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

3 0
3 years ago
A 6.00 cm tall light bulb is placed a distance of 54.2 cm from a double convex lens
BigorU [14]

Answer:

1 / f = 1 / i + 1 / o       thin lens equation

1 / i = 1 / f - 1 / o    =    (o - f) / (o * f)

i = o * f / (o - f)

i = 54.2 * 12.7 / (54.2 - 12.7) = 16.6 cm    image distance

Image is real and inverted and 16.6 / 54.2 * 6 = 1.94 cm tall

7 0
3 years ago
It takes 500 J of work to compress a spring 10 cm. What is the force constant of the spring? (b) When a 3.0 kg block is pushed a
NARA [144]

Answer:

Explanation:

a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²

500 = 5 x 10⁻³ k ,

k = (500/5) x 10³ = 10⁵ N/m

b )

k = 4.5 x 10¹ = 45 N/m

Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J

This energy gets dissipated by friction .

work done by friction = μ mg d

d is the distance traveled under friction

so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2

μ = 245 x 10⁻⁴  or 0.00245 which appears to be very small. .

8 0
3 years ago
Question about generators and electrical currents. Please, no absurd answers! Thank you
Kamila [148]
The first answer should be correct if not then the second one
3 0
3 years ago
Read 2 more answers
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