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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I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.
<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know
V and
I use formula
P = IV: P = IV = (100mA)(10V) = 1 W.</span>
The next question is what I'm not sure about:
Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).
What I did initially was: P = IV = (100mA)(2V) = 2 W
But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."
So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.
P = IV = (200mA)(2V) = 4 W
False, applied force is when a person or an object pushes on another object
Answer:

Explanation:
Given that,
Mass of the bowling ball, m = 5 kg
Radius of the ball, r = 11 cm = 0.11 m
Angular velocity with which the ball rolls, 
To find,
The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.
Solution,
The translational kinetic energy of the ball is :



The rotational kinetic energy of the ball is :



Ratio of translational to the rotational kinetic energy as :

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2