Ch 5.3 Help
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Ch 5.3 Help
Question 1: (1 point)
A bobsled at 158 km/hr turns around a curve with a 46.69 m radius of curvature on a banked ice track. What is the banking angle if banking alone (friction neglected) is to hold the sled on its track (see sheet 19,20)? Indicate with a negative (positive) sign whether the banking angle increases (decreases) with increasing speed.
(How to solve):
Question 2: (1 point)
A planet has a radius of 6.491 x106 m and a mass of 6.82 x1024 kg. What is the gravitational acceleration close to its surface (see sheet 21,22,25)? Indicate with a positive (negative) sign whether your calculated value depends (does not depend) on the chemical composition of the planet.
(How to solve):
Question 3: (1 point)
A rocket passes the earth at a large distance, of the order of earth radii. The gravitational acceleration due to the earth at this distance is 0.415 g ( g is the gravitational acceleration close to the surface of the earth; use 6.0x1024kg for the earth mass.) How far is the rocket from the center of the earth (see sheet27,27')? Indicate with a negative (positive) sign whether the gravitational acceleration varies strongly (hardly at all) with distance changes from the surface of the earth which are of the order of a few kilometers.
(How to solve):
Question 4: (1 point)
A satellite circles the earth at a height of 561.5 km above the surface of the earth. How many hours does it take the satellite to circle the earth (see sheet 31) (use 6.0x1024 kg and 6.4x106 m for the earth mass and radius)?
(How to solve):
A bobsled at 158 km/hr turns around a curve with a 46.69 m radius of curvature on a banked ice track. What is the banking angle if banking alone (friction neglected) is to hold the sled on its track (see sheet 19,20)? Indicate with a negative (positive) sign whether the banking angle increases (decreases) with increasing speed.
(How to solve):
Question 2: (1 point)
A planet has a radius of 6.491 x106 m and a mass of 6.82 x1024 kg. What is the gravitational acceleration close to its surface (see sheet 21,22,25)? Indicate with a positive (negative) sign whether your calculated value depends (does not depend) on the chemical composition of the planet.
(How to solve):
Question 3: (1 point)
A rocket passes the earth at a large distance, of the order of earth radii. The gravitational acceleration due to the earth at this distance is 0.415 g ( g is the gravitational acceleration close to the surface of the earth; use 6.0x1024kg for the earth mass.) How far is the rocket from the center of the earth (see sheet27,27')? Indicate with a negative (positive) sign whether the gravitational acceleration varies strongly (hardly at all) with distance changes from the surface of the earth which are of the order of a few kilometers.
(How to solve):
Question 4: (1 point)
A satellite circles the earth at a height of 561.5 km above the surface of the earth. How many hours does it take the satellite to circle the earth (see sheet 31) (use 6.0x1024 kg and 6.4x106 m for the earth mass and radius)?
(How to solve):
hwilson Guest
Question 2 and 3
Anyone have any idea on questions 2 and 3 have been working on them for a while and I think they are still due tomorrow. I cant figure it out please help?
gguest Guest
Question 1
Question 1: (1 point)
A bobsled at 158 km/hr turns around a curve with a 46.69 m radius of curvature on a banked ice track. What is the banking angle if banking alone (friction neglected) is to hold the sled on its track (see sheet 19,20)? Indicate with a negative (positive) sign whether the banking angle increases (decreases) with increasing speed.
(How to solve): (Answer is negative)
First convert the velocity they give to m/s.
158 km/hr * 1000 m/km = 158000m/hr / 3600 hr/s = 43.89 m/s
Now let's set up the equation. We know that the centripetal force is equal to the normal force. So mv^2/r = mgcosθ
The mass cancels out since it appears in both sides of the equation, so it's just v^2/r = gcosθ.
We know all of the variables in here now so just plug and solve for θ.
v^2/r = gcosθ
43.89^2/46.69 = 9.81cosθ
41.2558/9.81 = cosθ
4.205 = cosθ
1/4.205 = arccosθ
.2377 = arccosθ
arccos(.2377) = θ
θ = 76.24
A bobsled at 158 km/hr turns around a curve with a 46.69 m radius of curvature on a banked ice track. What is the banking angle if banking alone (friction neglected) is to hold the sled on its track (see sheet 19,20)? Indicate with a negative (positive) sign whether the banking angle increases (decreases) with increasing speed.
(How to solve): (Answer is negative)
First convert the velocity they give to m/s.
158 km/hr * 1000 m/km = 158000m/hr / 3600 hr/s = 43.89 m/s
Now let's set up the equation. We know that the centripetal force is equal to the normal force. So mv^2/r = mgcosθ
The mass cancels out since it appears in both sides of the equation, so it's just v^2/r = gcosθ.
We know all of the variables in here now so just plug and solve for θ.
v^2/r = gcosθ
43.89^2/46.69 = 9.81cosθ
41.2558/9.81 = cosθ
4.205 = cosθ
1/4.205 = arccosθ
.2377 = arccosθ
arccos(.2377) = θ
θ = 76.24
ravenscorne Posts : 15
Join date : 20080920
Question 2
Question 2: (1 point)
A planet has a radius of 6.491 x10^6 m and a mass of 6.82 x10^24 kg. What is the gravitational acceleration close to its surface (see sheet 21,22,25)? Indicate with a positive (negative) sign whether your calculated value depends (does not depend) on the chemical composition of the planet.
(How to solve): (Answer is negative)
To solve this problem, we use the gravitational acceleration equation, g= G*M/(d^2), where G is the universal gravitational constant, 6.7x10^(11), M is the mass of the planet (the object that an object gravitates to), and d is the distance from the center of the object, usually the radius.
g = G*M/(d^2)
g = (6.7x10^(11)*6.82x10^24)/((6.491x10^6)^2)
g = 10.845
A planet has a radius of 6.491 x10^6 m and a mass of 6.82 x10^24 kg. What is the gravitational acceleration close to its surface (see sheet 21,22,25)? Indicate with a positive (negative) sign whether your calculated value depends (does not depend) on the chemical composition of the planet.
(How to solve): (Answer is negative)
To solve this problem, we use the gravitational acceleration equation, g= G*M/(d^2), where G is the universal gravitational constant, 6.7x10^(11), M is the mass of the planet (the object that an object gravitates to), and d is the distance from the center of the object, usually the radius.
g = G*M/(d^2)
g = (6.7x10^(11)*6.82x10^24)/((6.491x10^6)^2)
g = 10.845
ravenscorne Posts : 15
Join date : 20080920
Question 3
Question 3: (1 point)
A rocket passes the earth at a large distance, of the order of earth radii. The gravitational acceleration due to the earth at this distance is 0.415 g ( g is the gravitational acceleration close to the surface of the earth; use 6.0x1024kg for the earth mass.) How far is the rocket from the center of the earth (see sheet27,27')? Indicate with a negative (positive) sign whether the gravitational acceleration varies strongly (hardly at all) with distance changes from the surface of the earth which are of the order of a few kilometers.
(How to solve): (Answer is positive)
In this problem, we use the same equation, but instead of solving for g, which we have, we solve for d.
Rearranging the equation, we get d = sqrt(G*M/g)
d = sqrt(G*M/g)
d = sqrt((6.7x10^(11)*6.0x10^24)/(.415*9.81))
d = 9936981.345
A rocket passes the earth at a large distance, of the order of earth radii. The gravitational acceleration due to the earth at this distance is 0.415 g ( g is the gravitational acceleration close to the surface of the earth; use 6.0x1024kg for the earth mass.) How far is the rocket from the center of the earth (see sheet27,27')? Indicate with a negative (positive) sign whether the gravitational acceleration varies strongly (hardly at all) with distance changes from the surface of the earth which are of the order of a few kilometers.
(How to solve): (Answer is positive)
In this problem, we use the same equation, but instead of solving for g, which we have, we solve for d.
Rearranging the equation, we get d = sqrt(G*M/g)
d = sqrt(G*M/g)
d = sqrt((6.7x10^(11)*6.0x10^24)/(.415*9.81))
d = 9936981.345
ravenscorne Posts : 15
Join date : 20080920
Question 4
Question 4: (1 point)
A satellite circles the earth at a height of 561.5 km above the surface of the earth. How many hours does it take the satellite to circle the earth (see sheet 31) (use 6.0x10^24 kg and 6.4x10^6 m for the earth mass and radius)?
(How to solve): (Answer is positive)
This question is the implementation of several equations. First we need to get the acceleration of satelite. From there, we can solve for the tangential velocity of the satelite, since we assume it is moving at a constant velocity. And finally we need to calculate the time. So to start off first, we use:
g = G*M/d^2 (Note: d in this instance needs to be modified. We can't use the radius of the earth because the satelite is rotating above earth. So we need to add the additional height to the radius of the earth to get the effective radius.)
d = 6.4x10^6 + .5615x10^6 = 6.9615x10^6
g = (6.7x10^(11)*6.0x10^24)/((6.9615x10^6)^2)
g = 8.2951
This is the gravitational acceleration, but if we think about it, it is also the centripetal acceleration of the satelite. So using a_c = v^2/r, we can solve for the tangential velocity. Rearranging this equation, we get:
v = sqrt(a_c/r) Where a_c = g
v = sqrt(8.2951*6.9615x10^6)
v = 7599.0904
Now we have the velocity, but we need the distance before we can solve for the time. We have the radius, so we can calculate the distance of the orbit, using the cirumference formula,
C = 2(pi)r
C = 2(3.14)(6.9615x10^6)
C = 43740394.52
Now we finally solve for time, using v = d/t
t = d/v
t = 43740394.52/7599.0904
t = 5756.003 seconds (Now convert to hours)
t = 5756.003/3600
t = 1.599 hours
A satellite circles the earth at a height of 561.5 km above the surface of the earth. How many hours does it take the satellite to circle the earth (see sheet 31) (use 6.0x10^24 kg and 6.4x10^6 m for the earth mass and radius)?
(How to solve): (Answer is positive)
This question is the implementation of several equations. First we need to get the acceleration of satelite. From there, we can solve for the tangential velocity of the satelite, since we assume it is moving at a constant velocity. And finally we need to calculate the time. So to start off first, we use:
g = G*M/d^2 (Note: d in this instance needs to be modified. We can't use the radius of the earth because the satelite is rotating above earth. So we need to add the additional height to the radius of the earth to get the effective radius.)
d = 6.4x10^6 + .5615x10^6 = 6.9615x10^6
g = (6.7x10^(11)*6.0x10^24)/((6.9615x10^6)^2)
g = 8.2951
This is the gravitational acceleration, but if we think about it, it is also the centripetal acceleration of the satelite. So using a_c = v^2/r, we can solve for the tangential velocity. Rearranging this equation, we get:
v = sqrt(a_c/r) Where a_c = g
v = sqrt(8.2951*6.9615x10^6)
v = 7599.0904
Now we have the velocity, but we need the distance before we can solve for the time. We have the radius, so we can calculate the distance of the orbit, using the cirumference formula,
C = 2(pi)r
C = 2(3.14)(6.9615x10^6)
C = 43740394.52
Now we finally solve for time, using v = d/t
t = d/v
t = 43740394.52/7599.0904
t = 5756.003 seconds (Now convert to hours)
t = 5756.003/3600
t = 1.599 hours
ravenscorne Posts : 15
Join date : 20080920
Ques 2, 3 & 4 Solutions...
I tried the solutions you posted for questions 2, 3 & 4 and my answers came up incorrect on Maple TA. I went back and retried, I checked for negative vs. positive signs, but it still didnt help...
Anyone have a clue as to how to go about this? Thanks so much.
Anyone have a clue as to how to go about this? Thanks so much.
PHYSTUDE Guest
Re: Ch 5.3 Help
Watch your inputs in the calculator. I can only guess that it is somewhere around entering 6.4x10^{24} (as an example only) that is messing up the calculations. If you are using a TI83 calculator I suggest using parenthesis and breaking down each part of the equation. I also suggest using the symbol "e" by hitting "2^{nd}" and then the comma button ",". Above it is is labeled "EE". You will see something like 6.4E^{24}. This is the same as saying "x10"
Guest01 Posts : 133
Join date : 20080919
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