Hovercraft B/C

[Justin72835]

A x directed force is applied to an object with the equation F_x = 5+.23x N. Find the work done by the force as the object moves from (0,0) to (7,0).

Answer (Click to Show Hidden Text)

Awesome calc question!

[math]W=\int_{0}^{7}Fdx=\int_{0}^{7}(5+.23x)dx=40.64 N[/math]

Edit: RIP, too slow; got ninja'd XD.

[Riptide]

A x directed force is applied to an object with the equation F_x = 5+.23x N. Find the work done by the force as the object moves from (0,0) to (7,0).

Answer (Click to Show Hidden Text)

Assuming coordinate system is in meters, [math]\int^{7 m}_{0 m} (5 + .23x) N dx = 40.635 J[/math]?

Sorry for not including units, nice job!

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

All right, next question! Ignore the completely unrealistic values.
An 80.0 kg astronaut experiences 0.100 'g's and throws a wrench that weighs 10.00 pound (at sea level on Earth) with an angle of elevation above his center of mass of 30.0 degrees. The wrench flies away from his hand at a speed of 100.0 m/s (assume that gravity does not affect the wrench for some reason).
a) How fast does he move after one second? Two seconds? Assume that gravity pulls on him downward with a constant force and that he will never hit ground. Also assume that the time spent throwing the wrench is negligible.
b) What is the impulse of the wrench on the astronaut? Assume constant acceleration on the wrench.
c) Suppose the mass of the planet the astronaut is above is 1.00 * 10^23 kg. How high does the astronaut need to be above the center of mass of the planet to experience the gravity he experiences?

[Justin72835]

All right, next question! Ignore the completely unrealistic values.
An 80.0 kg astronaut experiences 0.100 'g's and throws a wrench that weighs 10.00 pound (at sea level on Earth) with an angle of elevation above his center of mass of 30.0 degrees. The wrench flies away from his hand at a speed of 100.0 m/s.
a) How fast does he move after one second? Two seconds? Assume that gravity pulls on him downward with a constant force and that he will never hit ground. Also assume that the time spent throwing the wrench is negligible.
b) What is the impulse of the wrench on the astronaut? Assume constant acceleration on the wrench.
c) Suppose the mass of the planet the astronaut is above is 1.00 * 10^23 kg. How high does the astronaut need to be above the center of mass of the planet to experience the gravity he experiences?

Answer (Click to Show Hidden Text)

Pretty fun problem tbh.

A) 10 lb = 4.54 kg. Using conservation of momentum, we find that the speed of the astronaut immediately after throwing the wrench is 5.67 m/s. This velocity is divided up into two components, x and y: vx = 4.91 m/s and vy = 2.83 m/s. 

After one second, vy increases by 0.98 m/s, so using Pythagorean Theorem, we find that his total velocity is [b]6.21 m/s[/b].

After two seconds, vy increases by 1.96 m/s, so using the same procedure, we find that his total velocity is [b]6.86 m/s[/b].

B) Impulse is just change in momentum, which in this case is [b]452.6 kg m/s[/b].

C) a = Gm/r^2. Substituting a for 0.98 m/s^2 and m for 1e23 kg, the radius comes out to be [b]2610 km[/b].

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

All right, next question! Ignore the completely unrealistic values.
An 80.0 kg astronaut experiences 0.100 'g's and throws a wrench that weighs 10.00 pound (at sea level on Earth) with an angle of elevation above his center of mass of 30.0 degrees. The wrench flies away from his hand at a speed of 100.0 m/s.
a) How fast does he move after one second? Two seconds? Assume that gravity pulls on him downward with a constant force and that he will never hit ground. Also assume that the time spent throwing the wrench is negligible.
b) What is the impulse of the wrench on the astronaut? Assume constant acceleration on the wrench.
c) Suppose the mass of the planet the astronaut is above is 1.00 * 10^23 kg. How high does the astronaut need to be above the center of mass of the planet to experience the gravity he experiences?

Answer (Click to Show Hidden Text)

Pretty fun problem tbh.

A) 10 lb = 4.54 kg. Using conservation of momentum, we find that the speed of the astronaut immediately after throwing the wrench is 5.67 m/s. This velocity is divided up into two components, x and y: vx = 4.91 m/s and vy = 2.83 m/s. 

After one second, vy increases by 0.98 m/s, so using Pythagorean Theorem, we find that his total velocity is [b]6.21 m/s[/b].

After two seconds, vy increases by 1.96 m/s, so using the same procedure, we find that his total velocity is [b]6.86 m/s[/b].

B) Impulse is just change in momentum, which in this case is [b]452.6 kg m/s[/b].

C) a = Gm/r^2. Substituting a for 0.98 m/s^2 and m for 1e23 kg, the radius comes out to be [b]2601 km[/b].

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