Hovercraft B/C

[Justin72835]

Yep, although I have slightly different results from rounding. Your turn! (Also, 3 sigfigs for C )

A pipe system is constructed to allow the incompressible flow of water. At point A, the water has a velocity of 4.0 m/s and a pressure of 100 kPa and is flowing through a pipe of radius 0.8 m. The pipe then shifts upward such that its midpoint has risen 3.6 m and its radius has decreased to 0.6 m.

A) What is the volumetric flow rate of the water, which has units of m^3/s?

B) What is the velocity of the water at point B?

C) What is the pressure of the water at point B?

[Adi1008]

Yep, although I have slightly different results from rounding. Your turn! (Also, 3 sigfigs for C )

A pipe system is constructed to allow the incompressible flow of water. At point A, the water has a velocity of 4.0 m/s and a pressure of 100 kPa and is flowing through a pipe of radius 0.8 m. The pipe then shifts upward such that its midpoint has risen 3.6 m and its radius has decreased to 0.6 m.

A) What is the volumetric flow rate of the water, which has units of m^3/s?

B) What is the velocity of the water at point B?

C) What is the pressure of the water at point B?

Answer (Click to Show Hidden Text)

a) 8.04 m^3/s
b) 7.11 m/s
c) 47.443 kPa

[Justin72835]

Yep, although I have slightly different results from rounding. Your turn! (Also, 3 sigfigs for C )

A pipe system is constructed to allow the incompressible flow of water. At point A, the water has a velocity of 4.0 m/s and a pressure of 100 kPa and is flowing through a pipe of radius 0.8 m. The pipe then shifts upward such that its midpoint has risen 3.6 m and its radius has decreased to 0.6 m.

A) What is the volumetric flow rate of the water, which has units of m^3/s?

B) What is the velocity of the water at point B?

C) What is the pressure of the water at point B?

Answer (Click to Show Hidden Text)

a) 8.04 m^3/s
b) 7.11 m/s
c) 47.443 kPa

Perfect! You're next!

[Adi1008]

A pipe system is constructed to allow the incompressible flow of water. At point A, the water has a velocity of 4.0 m/s and a pressure of 100 kPa and is flowing through a pipe of radius 0.8 m. The pipe then shifts upward such that its midpoint has risen 3.6 m and its radius has decreased to 0.6 m.

A) What is the volumetric flow rate of the water, which has units of m^3/s?

B) What is the velocity of the water at point B?

C) What is the pressure of the water at point B?

Answer (Click to Show Hidden Text)

a) 8.04 m^3/s
b) 7.11 m/s
c) 47.443 kPa

Perfect! You're next!

A bullet of mass m and velocity v is fired towards a block of mass 5m. The block is initially at rest on a frictionless surface. The bullet enters the block and exits with 1/4th of its original kinetic energy.

a) What is the final speed of the block?
b) What is the final speed of the bullet?
c) What is the gain in the kinetic energy of the block?

[UTF-8 U+6211 U+662F]

A bullet of mass m and velocity v is fired towards a block of mass 5m. The block is initially at rest on a frictionless surface. The bullet enters the block and exits with 1/4th of its original kinetic energy.

a) What is the final speed of the block?
b) What is the final speed of the bullet?
c) What is the gain in the kinetic energy of the block?

Answer (Click to Show Hidden Text)

a)

[math]\frac12mv^2 = 4\left(\frac12mv_f^2\right)[/math]

[math]\frac{v^2}{4} = v_f^2[/math]

[math]\frac{v}{2} = v_f[/math]

[math]mv = (5m)v_{block} + m\left(\frac{v}{2}\right)[/math]

[math]v_{block} = \frac{mv - \frac{mv}{2}}{5m}[/math]

[math]v_{block} = \frac{mv}{10m} = \frac{v}{10}[/math]

b)

[math]v_{bullet_f} = \frac{v}{2}[/math]

c)

[math]KE_{block} = \frac12(5m)v_{block}^2 = \frac{5m*\frac{v^2}{100}}{2}[/math]

[math]KE_{block} = \frac{mv^2}{40}[/math]

[math]\Delta KE_{block} = \frac{mv^2}{40} - 0 = \frac{mv^2}{40}[/math]

[Adi1008]

A bullet of mass m and velocity v is fired towards a block of mass 5m. The block is initially at rest on a frictionless surface. The bullet enters the block and exits with 1/4th of its original kinetic energy.

a) What is the final speed of the block?
b) What is the final speed of the bullet?
c) What is the gain in the kinetic energy of the block?

Answer (Click to Show Hidden Text)

a)

[math]\frac12mv^2 = 4\left(\frac12mv_f^2\right)[/math]

[math]\frac{v^2}{4} = v_f^2[/math]

[math]\frac{v}{2} = v_f[/math]

[math]mv = (5m)v_{block} + m\left(\frac{v}{2}\right)[/math]

[math]v_{block} = \frac{mv - \frac{mv}{2}}{5m}[/math]

[math]v_{block} = \frac{mv}{10m} = \frac{v}{10}[/math]

b)

[math]v_{bullet_f} = \frac{v}{2}[/math]

c)

[math]KE_{block} = \frac12(5m)v_{block}^2 = \frac{5m*\frac{v^2}{100}}{2}[/math]

[math]KE_{block} = \frac{mv^2}{40}[/math]

[math]\Delta KE_{block} = \frac{mv^2}{40} - 0 = \frac{mv^2}{40}[/math]

Looks good to me; your turn!

[UTF-8 U+6211 U+662F]

Inclined planes are a simple machine, of which hills are a type.
1) Explain why climbing a steeper hill is harder than climbing a shallower hill.
2) Explain why the work done pushing a ball up a hill is the same no matter the angle of the inclined plane if the height is the same.
3) Explain why a person can walk up the hill multiple times and do different amounts of work.

[Tesel]

Inclined planes are a simple machine, of which hills are a type.
1) Explain why climbing a steeper hill is harder than climbing a shallower hill.
2) Explain why the work done pushing a ball up a hill is the same no matter the angle of the inclined plane if the height is the same.
3) Explain why a person can walk up the hill multiple times and do different amounts of work.

Answer (Click to Show Hidden Text)

1) Inclined planes are simple machines, so they decrease the input force needed by increasing distance. A steeper hill has smaller distance, so the input force has to be higher. This is because of the relationship W = Fd.
2) The work is based on change in GPE. This can be calculated from GPE = mgh. Thus, only the height will matter for change in GPE, not horizontal distance.
3) Walking up a hill implies that there is friction between the person and the hill. Unlike GPE, friction is pathway-dependent, so the person could have to put varying amounts of work in as energy is converted to heat and sound through friction.

[UTF-8 U+6211 U+662F]

Inclined planes are a simple machine, of which hills are a type.
1) Explain why climbing a steeper hill is harder than climbing a shallower hill.
2) Explain why the work done pushing a ball up a hill is the same no matter the angle of the inclined plane if the height is the same.
3) Explain why a person can walk up the hill multiple times and do different amounts of work.

Answer (Click to Show Hidden Text)

1) Inclined planes are simple machines, so they decrease the input force needed by increasing distance. A steeper hill has smaller distance, so the input force has to be higher. This is because of the relationship W = Fd.
2) The work is based on change in GPE. This can be calculated from GPE = mgh. Thus, only the height will matter for change in GPE, not horizontal distance.
3) Walking up a hill implies that there is friction between the person and the hill. Unlike GPE, friction is pathway-dependent, so the person could have to put varying amounts of work in as energy is converted to heat and sound through friction.

Yep, your turn!

[Tesel]

I owe you guys a question, sorry about that... if anyone has a question they'd like to ask, feel free to post it, otherwise I'll type one up tonight.