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shtirl [24]
3 years ago
5

The caste system is an example of​

Physics
1 answer:
snow_lady [41]3 years ago
7 0

Answer:

It is example of Closed system

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Imagine an aluminum cup of 0.30 liter capacity filled with mercury at 14.0 degrees C. How much mercury, if any, will spill out o
tresset_1 [31]

Answer:

ΔV=0.00043 L

Explanation:

Given that

V= 0.3 L

T= 14 °C

T'=27°C

Coefficient of volume expansion

\beta _{mercury}=1.8\times 10^{-4}\ C^{-1}

\beta _{Almunium}=0.69\times 10^{-4}\ C^{-1}

\Delta V=(\beta _{mercury}-\beta _{Almunium} )V\Delta T

\Delta V=(1.8\times 10^{-4} - 0.69\times 10^{-4})\times 0.3\times (27-14)

ΔV=0.00043 L

4 0
3 years ago
Which of the following is an explanation of how the natural world works, based on experimentation?
Nostrana [21]

Answer:

It is an analysis.....

5 0
4 years ago
Read 2 more answers
Una onda sonora se produce durante 1,5 s. Posee una longitud de onda de 2,4 m y una velocidad de 340 m/s.
cestrela7 [59]

Answer:

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

Explanation:

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

3 0
3 years ago
A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha
AlexFokin [52]

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as

m_1u_1+m_2u_2 = (m_1+m_2)v_f

Here

m_1 = Mass of big fish

m_2 = Mass of small fish

v_1 = Velocity of big fish

v_2 = Velocity of small fish

v_F = Final Velocity

The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have

m_1u_1=-m_2u_2

u_2 = -\frac{m_1u_1}{m_2}

Replacing we have,

u_2 = -\frac{(5kg)(1m/s)}{0.5kg}

u_2 = -10m/s

The negative sign indicates that the small fish is swimming in the direction opposite to that of the big fish.

Therefore the speed of the small fish is 10m/s

8 0
3 years ago
A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W
Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                             V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)

                             V^{2} = 24.5 m/s

                             V = \sqrt{24.5} \ m/s

                             V = 4.95 \ m/s

                            V = ± 4.95 m/s

                            V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is  V = - 4.95 m/s  

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s  

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                            0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)

                            0 = U^{2} - 16.072 m/s

                            U^{2} = 16.072 m/s

                            U = \sqrt{16.072} \ m/s

                           U = ± 4.01 m/s

                          U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is  U = + 4.01 m/s                        

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

                            F = \frac{mv - mu}{t}

                          F.t = m(v - u)

       ⇒      Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,                        

          ∴          F.t = 0.120  kg(4.01  m/s - (-4.95  m/s) )

                      F.t = 0.120  kg(4.01  m/s + 4.95  m/s) )

                      F.t = 0.120  kg × 8.96  m/s

                      Impulse  = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

                             

6 0
3 years ago
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