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

1) For the equation delta x = ct^3, the SI units of the proportionality constant c can be determined by analyzing the dimensions of both sides of the equation.

The dimensions of delta x are in meters (m) since it represents displacement. The dimensions of t are in seconds (s) since it represents time. Therefore, the dimensions of ct^3 would be (m/s^3) * (s^3) = m.

Therefore, the SI units of the proportionality constant c would be meters per second cubed (m/s^3).

2) Let's analyze the dimensions of the equation K = 1/2mv^2.

The dimensions of K are in Joules (J) since it represents kinetic energy. The dimensions of m are in kilograms (kg) since it represents mass. The dimensions of v are in meters per second (m/s) since it represents velocity.

Therefore, the dimensions of 1/2mv^2 would be (kg) * (m/s)^2 = kg * (m^2/s^2).

Comparing the dimensions of the left and right sides of the equation, we see that they are both in kg * (m^2/s^2), so the dimensions are correct. The equation is dimensionally correct.

3) To find the volume (V) of the Earth, we can use the given average density (D) and mass (M).

V = M/D

Given:
Mass of the Earth (M) = 5.98E24 kg
Average density of the Earth (D) = 5.520 kg/m^3

By substituting the values into the formula, we get:

V = (5.98E24 kg) / (5.520 kg/m^3)

Calculating this expression gives us the volume of the Earth in cubic meters (m^3).

To find the radius (r) of the Earth, we can use the formula for the volume of a sphere:

V = (4/3)πr^3

Rearranging this equation, we have:

r = [(3V) / (4π)]^(1/3)

Substituting the calculated value of V into this formula, we can find the radius of the Earth in meters (m).

To convert the radius from meters to miles, we can divide the radius in meters by the conversion factor of 1609.3 meters/mile. This will give us the radius of the Earth in miles (mi).