It is B because the other ones are good.
Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.
- 1. Ball A will have the greater density
- 2. Ball C and Ball D have the same density.
- 3. Ball Q will have the greater density.
- 4. Ball X and Y will have the same density
The density of an object is given as its mass per unit volume of the object.
Mathematically;.
For Case 1:
- Va = Vb and Ma = 2Mb
- D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
- Therefore, the density of ball A,
- D(a) = 2D(b).
- Therefore, ball A has the greater density.
For Case 2:
- D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd
- Therefore, ball C and D have the same density
For Case 3:
- Vp = 2Vq and Mp = Mq
- D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
- Therefore, the density of ball P is half the density of ball Q
- Therefore, ball Q has the greater density.
For case 4:
Therefore, Ball X and Ball Y have the same density.
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brainly.com/question/18110802
You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).
And you know that the potential energy, PE, is [ 1/2 ] k (x^2)
Then, use x = A, to calculate the PE in the point where ME = PE.
ME = PE = [1/2] k (A)^2.
At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2
=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME
So, if PE is 1/4 of ME, KE is 3/4 of ME.
And the answer is 3/4
You pick a system for which no control sample exists, so that no one can show that the alleged causal relationships you assert do not, in fact, lead to the phenomenon you claim to have observed.
