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4 years ago

14

Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. If the incoming ball has a speed of 20

0 m/s . Part APart complete What is the final speed of the incoming ball if it is much more massive than the stationary ball?

2 answers:

Bingel [31]4 years ago

8 0

Answer:

Explanation:

Check the attachment for solution

andrew11 [14]4 years ago

8 0

Answer: The final speed of the incoming ball is approximately 200m/s

Explanation:

Using the law of conservation of momentum.

m1(u1) + m2u2 = m1v1 + m2v2

And also law of conservation of kinetic energy for elastic heads on collision we can derive the formula for elastic heads on collision which is given below:

For elastic heads on collision.

v1 = [( m1 - m2)/(m1+m2)] u1 ......1

v2 = [(2m1)/(m1+m2)]u1 ......2

Where,

m1 and m2 are the mass of the incoming and stationary ball respectively.

u1 and u2 are the initial speed of the incoming and stationary ball respectively.

v1 and v2 are the final speed of the incoming and stationary ball respectively.

a) to determine the final speed of the incoming ball using equation 1

v1 = [( m1 - m2)/(m1+m2)]u1

Since m1 >> m2

m1 - m2 ~= m1 and m1 +m2 ~= m1

So, equation 1 becomes

v1 ~= [m1/m1]u1

v1 ~= u1

Since u1 = 200m/s

v1 ~= 200m/s

Additional tips: using equation 2 we can derive the approximate final speed of the stationary ball following the same assumptions. If well solved v2 = 2u1 = 400m/s

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defon

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A man with a mass of 70 kg walks up a flight of stairs that is 5 m tall. Another man with a mass of 80 kg walks up the same flig

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3 years ago

Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.

Vesnalui [34]

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

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4 years ago

A copper wire has a square cross section 2.0 mm on a side. The wire is 5.0 m long and carries a current of 2.0 A. The density of

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Answer:

30.22 hours

Explanation:

Given data:

A= l² = (2 x $10^{-3}$ )² = 4 x $10^{-6}$ m²

Length 'L' = 5m

current ' $I$ ' = 2 A

density of free electrons 'n'= 8.5 x $10^{28}$ /m³

Current Density 'J' = $I$ / A

J= 2/4 x $10^{-6}$

J= 5 x $10^{5}$ A/m²

We can determine the time required for an electron to travel the length of the wire by

T= L/ Vd

Where,

L is length and Vd is drift velocity.

Vd can be defined by J/ n|q|

where,

n is the charge-carrier number density

|q| is is the charge carried by each charge carrier =>1.6 x $10^{-19}$ C

T= L/ Vd

Therefore,

T= L . n|q| / J

T= (4 x 8.5 x $10^{28}$ x |1.6 x $10^{-19}$ |)/5 x $10^{5}$

T= 108800 seconds =>1813.33 minutes

Converting minute into hours:

T= 30.22 hours

Thus, time that is required for an electron to travel the length of the wire is 30.22 hours

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3 years ago

A que profundidad esta nadando una persona dentro de una alberca si la presión absoluta sobre ésta es de 156kPa?

lara31 [8.8K]

Answer:

La persona está nadando en la alberca a una profundida de 5.575 metros.

Explanation:

La presión absoluta ( $P_{tot}$ ) experimentada por la persona es la suma de la presión atmosférica ( $P_{atm}$ ) y la presión hidrostática de la columna de agua de la alberca ( $P_{h}$ ), medidas en kilopascales. Es decir,

$P_{tot} = P_{atm}+P_{h}$ (1)

$P_{tot} = P_{atm} + \frac{\rho\cdot g \cdot z}{1000}$ (2)

Donde:

$\rho$ - Densidad del fluido de la alberca, medida en kilogramos por metro cúbico.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$z$ - Profundidad de la persona en la alberca, medida en metros.

Si sabemos que $P_{atm} = 101.325\,kPa$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ y $P_{tot} = 156\,kPa$ , entonces la profundidad de la persona en la alberca es:

$156 = 101.325 +\frac{(1000)\cdot (9.807)\cdot z}{1000}$

$54.675 = 9.807\cdot z$

$z = 5.575\,m$

La persona está nadando en la alberca a una profundida de 5.575 metros.

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3 years ago