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gayaneshka [121]
3 years ago
6

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

with an initial velocity of 3.00j. Find magnitude and direction of the final velocity of the composite.
Physics
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

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g An electric hot plate raises its own internal energy and the internal energy of a cup of water by 9000J, and there is at the s
DochEvi [55]

Answer:

11700j

Explanation:

add the two because the plate has to maintain the temp.

2700+9000=11700

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3 years ago
A garrafa térmica (também conhecida como "vaso de Dewar") é um dispositivo extremamente útil para conservar, no seu interior, co
igor_vitrenko [27]

Answer:

A opção A está correta.

O sistema formado pela garrafa térmica e a água perde 400 cal de calor para o meio ambiente.

Option A is correct.

The system formed by the thermos and the water loses 400 cal of heat to the environment.

Explanation:

Quando a temperatura de um sistema reduz, fica claro que o sistema perdeu calor ou energia térmica. Como a temperatura é um dos indicadores mais claros disso, esta conclusão é hermética e correta.

Mas, para saber a quantidade de calor perdida para o meio ambiente, agora fazemos alguns cálculos de energia térmica.

Transferência de calor de ou para o sistema de água e garrafa térmica = c × ΔT

c = capacidade térmica do sistema de água e garrafa térmica = 80 cal /°C

ΔT = Alteração da temperatura do sistema de água e garrafa térmica = (temperatura final) - (temperatura inicial) = 55 - 60 = -5°C

Calor transferido = 80 × -5 = -400 cal.

O sinal de menos mostra que o calor é transferido para fora do sistema, ou seja, o calor é perdido no sistema.

Espero que isto ajude!!!

English Translation

The thermos (also known as "Dewar vase") is an extremely useful device to conserve bodies (essentially liquid) at high temperatures, minimizing energy exchanges with the environment, which is generally colder. A thermos contains water at 60 o C. The thermos + water set has a thermal capacity of C = 80 cal / o C. The system is placed on a table and, after a considerable period of time, its temperature decreases to 55 o C. In this case, it is concluded that the system formed by the thermos and the water inside:

a) lost 400 cal. B) gained 404cal. C) lost 4 850 cal. D) gained 4 850 cal. E) did not exchange heat with the external environment.

Solution

When a system's temperature reduces, it is clear to conclude that the system has lost heat or thermal energy. Since temperature is one of clearest indicators of this, this conclusion is airtight and correct.

But, to know the amount of heat lost to the environment, we now do some thermal energy calculations.

Heat transferrred from or to the water and thermos system = c × ΔT

c = heat capacity of the water and thermos system = 80 cal/°C

ΔT = Change in temperature of the water and thermos system = (final temperature) - (initial temperature)

= 55 - 60 = -5°C

Heat transferred = 80 × -5 = -400 cal.

The minus sign shows that the heat is transferred out of the system, that is, the heat is lost from the system.

Hope this Helps!!!

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Answer:

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

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Explanation:

In the measurements of the oscillatory movement in general, the most difficult magnitude of the measurement is the period, by reasoning

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

* Problems to synchronize when starting to measure the oscillation and end point of the oscilloscope, this error needs a good control system to be decreased.

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