Ch 13.1 help
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Ch 13.1 help
Question 1
A constant force of 121.5 N acts tangentially on the wheel shown which has a 41.22 cm radius. The wheel axis through the center point is inserted into a hole where it has a tight fit causing friction which generates heat. How much heat in calories is generated during 290.5 turns (see sheet 3; calculate the work done by the force against the frictional torque see sheet Ch 8 sheet 38) ?
Answer: 21832.693079432847
How to solve:
Question 2
A 30.83 kg person runs 43.22 times up the stairs to the D level of the Physics Building. Assume for simplicity that the energy spent is equal to the gain in potential energy for the 29.22 m height. How many nutritional calories has the person used up (see sheet 3) ? Indicate with a positive (negative) sign whether under your simplifying assumption it does (does not) matter if the person walks or runs the distance.
Answer: 91.2230409217387
How to solve:
Question 3
A bridge made of steel and 0.7772 km long undergoes temperature changes from 6.063oF to 110.42oF. By how much does the length of the bridge change due to thermal expansion (see sheet 4; use 11x106 oC1 for the coefficient of linear expansion of steel) ? Indicate with a positive (negative) sign whether the length increases ( decreases ) with rising temperature.
Answer: 0.49564936911111107
How to solve:
Question 4
The temperature of 3.182 liters of water is raised by 44.03 oC. What is the heat input in calories (see sheet 8,11) ? Indicate with a negative (positive) sign whether for the same mass in steel less (more) calories are needed to produce the same temperature rise (use 0.107 cal g1 oC1 for the specific heat of steel).
Answer: 140103.46
How to solve:
A constant force of 121.5 N acts tangentially on the wheel shown which has a 41.22 cm radius. The wheel axis through the center point is inserted into a hole where it has a tight fit causing friction which generates heat. How much heat in calories is generated during 290.5 turns (see sheet 3; calculate the work done by the force against the frictional torque see sheet Ch 8 sheet 38) ?
Answer: 21832.693079432847
How to solve:
Question 2
A 30.83 kg person runs 43.22 times up the stairs to the D level of the Physics Building. Assume for simplicity that the energy spent is equal to the gain in potential energy for the 29.22 m height. How many nutritional calories has the person used up (see sheet 3) ? Indicate with a positive (negative) sign whether under your simplifying assumption it does (does not) matter if the person walks or runs the distance.
Answer: 91.2230409217387
How to solve:
Question 3
A bridge made of steel and 0.7772 km long undergoes temperature changes from 6.063oF to 110.42oF. By how much does the length of the bridge change due to thermal expansion (see sheet 4; use 11x106 oC1 for the coefficient of linear expansion of steel) ? Indicate with a positive (negative) sign whether the length increases ( decreases ) with rising temperature.
Answer: 0.49564936911111107
How to solve:
Question 4
The temperature of 3.182 liters of water is raised by 44.03 oC. What is the heat input in calories (see sheet 8,11) ? Indicate with a negative (positive) sign whether for the same mass in steel less (more) calories are needed to produce the same temperature rise (use 0.107 cal g1 oC1 for the specific heat of steel).
Answer: 140103.46
How to solve:
periwinkle Posts : 22
Join date : 20080917
Question 1
We are looking for the amount of Heat in calories. We can find it in joules easily and just need to do a conversion factor given to us on Ch 13, Sheet 3.
Q_{calories} = ω/4.187
ω = F*r
Let's find the amount of work first:
ω = 121.5 N * 0.4122 [Don't forget to convert it into S.I. Units!]
ω = 50.0823
Not exactly there yet...
Since it is a spinning disk and the heat is found after s number of turn we have to include this into the equation. Here it makes 290.5 and since it's circular you CANNOT forget to include 2π. Just multiply the rotation and the work:
Q_{joules} = ω*rotation
Q_{joules} = 50.0823*(290.5*2π)
Q_{joules} = 91413.48592
Remember! Calculate the rotation first then multiply it by the work! If you are still not getting this answer make sure you calculator is in RADIANS, these are two very important and overlooked details and can save a lot of "Why am I getting this wrong..?"
Now just convert it into calories:
Q_{calories} = 91413.48592/4.187
Q_{calories} =
Q_{calories} = ω/4.187
ω = F*r
Let's find the amount of work first:
ω = 121.5 N * 0.4122 [Don't forget to convert it into S.I. Units!]
ω = 50.0823
Not exactly there yet...
Since it is a spinning disk and the heat is found after s number of turn we have to include this into the equation. Here it makes 290.5 and since it's circular you CANNOT forget to include 2π. Just multiply the rotation and the work:
Q_{joules} = ω*rotation
Q_{joules} = 50.0823*(290.5*2π)
Q_{joules} = 91413.48592
Remember! Calculate the rotation first then multiply it by the work! If you are still not getting this answer make sure you calculator is in RADIANS, these are two very important and overlooked details and can save a lot of "Why am I getting this wrong..?"
Now just convert it into calories:
Q_{calories} = 91413.48592/4.187
Q_{calories} =
 Spoiler:
 21832.69308
Guest01 Posts : 133
Join date : 20080919
Question 3
This one is a bit more annoying because of the conversions, personally I don't like converting only because it's more work.
Watch S.I. Units! In fact let's just do all our converting now!
T_{1} = (°F_{1}32)*(5/9)
T_{1} = (110.42°F32)*(5/9)
T_{1} = 43.56666667T °C
T_{2} = (6.063°F32)*(5/9)
T_{2} = 14.40944444 °C
ΔT = (T_{1}T_{2})
ΔT = 57.97611111 °C
We can leave our answer in Celsius. ΔT would be the same for Kelvin
Now our length has to be in meters. It's given in kilometers and to convert to meters just multiply by 1000
L_{0} = 0.7772 km * 1000
L_{0} = 777.2 m
Now we can use the formula. The number for α is already given to us as 11x10^{6}, now plug in everything and use:
(ΔL/L_{0}) = α*ΔT
ΔL = L_{0}*α*ΔT
ΔL = 777.2 m * 11x10^{6} * 57.97611111 °C
ΔL =
Watch S.I. Units! In fact let's just do all our converting now!
T_{1} = (°F_{1}32)*(5/9)
T_{1} = (110.42°F32)*(5/9)
T_{1} = 43.56666667T °C
T_{2} = (6.063°F32)*(5/9)
T_{2} = 14.40944444 °C
ΔT = (T_{1}T_{2})
ΔT = 57.97611111 °C
We can leave our answer in Celsius. ΔT would be the same for Kelvin
Now our length has to be in meters. It's given in kilometers and to convert to meters just multiply by 1000
L_{0} = 0.7772 km * 1000
L_{0} = 777.2 m
Now we can use the formula. The number for α is already given to us as 11x10^{6}, now plug in everything and use:
(ΔL/L_{0}) = α*ΔT
ΔL = L_{0}*α*ΔT
ΔL = 777.2 m * 11x10^{6} * 57.97611111 °C
ΔL =
 Spoiler:
 0.4956493691
Guest01 Posts : 133
Join date : 20080919
Question 4
This is real easy, the hardest part is realizing you have to convert Liters to weight. Instead of using all math I already know that the volume of water in Liters can be converted to mass in grams by multiplying it by 1000. You can double check this by converting Liters (volume) to cubic centimeters and since 1g of water is equal to a cubic centimeter we can multiply it by that. Actually I'm not 100 percent sure if those are the correct S.I. Units trough the conversions. You should really double check this. However it is the same concept and procedure to convert it.
ΔT and c are given to us already. I ignored c since I know the value is just 1 and can be ignored. Don't do this! This is bad practice by me and understand you have to include c.
ΔT= 44.03
c = 1
Now using the formula multiply it out:
Q = m*c*ΔT
Q = (3.182*1000)*(1)*44.03
Q =
ΔT and c are given to us already. I ignored c since I know the value is just 1 and can be ignored. Don't do this! This is bad practice by me and understand you have to include c.
ΔT= 44.03
c = 1
Now using the formula multiply it out:
Q = m*c*ΔT
Q = (3.182*1000)*(1)*44.03
Q =
 Spoiler:
 140103.46
Guest01 Posts : 133
Join date : 20080919
Question 2
Took me a bit of time to figure this one out, then I had a realization, we're supposed to treat as Potential Energy (completely overlooked it in the question).
First use the formula for PE and you should end up with a value close to 8837.364006
This is done a certain amount of times so include that:
8837.364006 * 43.22 = 381950.04092 J
Now convert the joules into calories the same way as we did for Question 1
381950.04092/4.187 = 91223.04092
Now it is asking for nutritional or kcal which is one thousandth of a calorie so just divide by 1000 (or multiply by 0.001)
91223.04092/1000 =
First use the formula for PE and you should end up with a value close to 8837.364006
This is done a certain amount of times so include that:
8837.364006 * 43.22 = 381950.04092 J
Now convert the joules into calories the same way as we did for Question 1
381950.04092/4.187 = 91223.04092
Now it is asking for nutritional or kcal which is one thousandth of a calorie so just divide by 1000 (or multiply by 0.001)
91223.04092/1000 =
 Spoiler:
 91.22304092
Guest01 Posts : 133
Join date : 20080919
Re: Ch 13.1 help
In 121.5 N? The N is just the units, that is N = Newtons; so it's 121.5 Newtons
Guest01 Posts : 133
Join date : 20080919
Re: Ch 13.1 help
I'm not quite sure what the (2n) is. What's n? when you multiply your rotation and your w together, what's the 2n? n must be some value because without it you wouldnt be getting the answer that you've calculated on number one. thanks for your help!
question Guest
Re: Ch 13.1 help
that's not supposed to be an "n" sorry, it's just the way the font is. It's actually supposed to be "pi" so you're multiplying by "2pi".
Yes it's not due until December, however I'm trying to get a lot of my work done early.
Yes it's not due until December, however I'm trying to get a lot of my work done early.
Guest01 Posts : 133
Join date : 20080919
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