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.
Read more:
brainly.com/question/18110802
Answer:
Imp = 25 [kg*m/s]
v₂= 20 [m/s]
Explanation:
In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.
1)
where:
m₁ = mass of the object = 5 [kg]
v₁ = initial velocity = 0 (initially at rest)
F = force = 5 [N]
t = time = 5 [s]
v₂ = velocity after the momentum [m/s]
![(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]](https://tex.z-dn.net/?f=%285%2A0%29%20%2B%285%2A5%29%20%3D%20%28m_%7B1%7D%2Av_%7B2%7D%29%20%3D%20Imp%5C%5CImp%20%3D%2025%20%5Bkg%2Am%2Fs%5D)
2)
![(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]](https://tex.z-dn.net/?f=%28m_%7B1%7D%2Av_%7B1%7D%29%2B%28F%2At%29%3D%28m_%7B1%7D%2Av_%7B2%7D%29%5C%5C%280.075%2A0%29%2B%2830%2A0.05%29%3D%280.075%2Av_%7B2%7D%29%5C%5Cv_%7B2%7D%3D20%20%5Bm%2Fs%5D)
Answer : 6.022• 10^23 atoms of potassium
