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alexdok [17]
4 years ago
7

A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to

eachother, at 18 m/s. The third piece has twice the mass asthe other two.Part 1 What is the speed of the third piece? (Answer in 2 sig figs andm/s)Part 2 What is the direction of the third piece? (Answer in 2 sig figs anddegrees north of east)
Physics
1 answer:
loris [4]4 years ago
6 0

Answer:

V = 13m/s

North-East (i.e. 45 degree from North)

Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of V_{x} (horizontal component of velocity) and V_{y} (vertical component of velocity)

We have

V_{y} = Velocity in South direction = 18m/s

m_{y} = Mass of piece in South direction = m

V_{x} = Velocity in West direction = 18m/s

m_{x} = Mass of piece in West direction = m

V_{ry} = Velocity of Third Piece in North direction = unknown

V_{rx} = Velocity of Third Piece in East direction = unknown

m = Mass of third piece = 2m

mV_{ry}  = m_{y} V_{y} \\ 2mV_{ry}  = m18\\ V_{ry} = 9

mV_{rx}  = m_{x} V_{x} \\ 2mV_{rx}  = m18\\ V_{rx} = 9

Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .

We can simply apply the Pythagoras theorem and find the hypotenuse

Hypotenuse^{2}  = Height^{2}  + Base^{2} \\ V_{3}^2  = 9^{2} + 9^{2} \\ V = 12.728 m/s

V = 13m/s

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La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.
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Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

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Usando la ecuación de movimiento

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2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

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b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

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Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

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