Asked by galileo
Suppose that the displacement of a particle is related to time according to the expression delta x = ct^3. What are the SI units of the proportionality constant c?
My answer: delta x = ct^3
m = c s^3 (s: seconds)
m/s^3= c
The SI units for c: m/s^3 correct?
2) Determine if dimensions are correct:
K = 1/2mv2
kgm^2/s^2 = kgm^2/s^2 * 1/2
kgm^2/s^2 = kgm^2/2s^2
The above equation is not dimensionally correct. However, I am somewhat confused as this is suppose to be the kinetic energy formula, and I just found that it is not dimensionally correct. Did I go wrong somewhere?
3) The mass of the Earth is 5.98E24 kg and its average density is 5.520kg/m^3. Assuming the Earth is spherical, find (a)its volume in m^3, (b) its radius in m, and (c) its radius in mi (miles).
My half solution:
Although I know the volume of a sphere is 4/3 pi r^3, here it throws me off.
I can find the volume simply by:
5,520kg = m^3/5.98E24kg; cross multiply and kg cancel(?) leaving me with V(earth)= 2.15E6 m^3(?)
Here, I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?
Thank you so much.
<<I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?
>>
Since you know V, rearrange that equation to get
r = (cube root of)[3 V/(4 pi)]
To convert meters to miles, divide by 1609.3 meters/mile
<<I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?
>>
Since you know V, rearrange that equation to get
r = (cube root of)[3 V/(4 pi)]
To convert meters to miles, divide by 1609.3 meters/mile
(1)Your answer is correct
(2) 1 kg m^2/s^2 is the same as 1 Joule, or 1 Newton-meter. The KE equation IS dimensionally correct
My answer: delta x = ct^3
m = c s^3 (s: seconds)
m/s^3= c
The SI units for c: m/s^3 correct?
2) Determine if dimensions are correct:
K = 1/2mv2
kgm^2/s^2 = kgm^2/s^2 * 1/2
kgm^2/s^2 = kgm^2/2s^2
The above equation is not dimensionally correct. However, I am somewhat confused as this is suppose to be the kinetic energy formula, and I just found that it is not dimensionally correct. Did I go wrong somewhere?
3) The mass of the Earth is 5.98E24 kg and its average density is 5.520kg/m^3. Assuming the Earth is spherical, find (a)its volume in m^3, (b) its radius in m, and (c) its radius in mi (miles).
My half solution:
Although I know the volume of a sphere is 4/3 pi r^3, here it throws me off.
I can find the volume simply by:
5,520kg = m^3/5.98E24kg; cross multiply and kg cancel(?) leaving me with V(earth)= 2.15E6 m^3(?)
Here, I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?
Thank you so much.
<<I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?
>>
Since you know V, rearrange that equation to get
r = (cube root of)[3 V/(4 pi)]
To convert meters to miles, divide by 1609.3 meters/mile
<<I'm lost about finding the radius in m and radius in miles. Should I use formula 4/3 pi r^3?
>>
Since you know V, rearrange that equation to get
r = (cube root of)[3 V/(4 pi)]
To convert meters to miles, divide by 1609.3 meters/mile
(1)Your answer is correct
(2) 1 kg m^2/s^2 is the same as 1 Joule, or 1 Newton-meter. The KE equation IS dimensionally correct
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