Answer:
The capacitance is cut in half.
Explanation:
The capacitance of a plate capacitor is directly proportional to the area A of the plates and inversely proportional to the distance between the plates d. So if the distance was doubled we should expect that the capacitance would be cut in half. That can be verified by the following equation that is used to compute the capacitance in such cases:
C = (\epsilon)*(A/d)
Where \epsilon is a constant that represents the characteristics for the insulator between the plates. A is the area of the plates and d is the distance between them. When we double d we have a new capacitance, given by:
C_new = (\epsilon)*(A/2d)
C_new = (1/2)*[(\epsilon)*(A/d)]
Since C = (\epsilon)*(A/d)] we have:
C_new = (1/2)*C
Answer:
The points 2 and 4 should be connected.
Explanation:
To complete the circuit, we need to connect the two points which when connected, encompass the battery and the bulb in the circuit. The points 2 and 4 do the job, since they connect the terminal of the battery and the terminal of the bulb, and thus complete the circuit.
Therefore, the choice C is correct.
Answer:
SI unit of k (spring constant) = N/m
Explanation:
We have expression for force in a spring extended by x m given by
F = kx
Where k is the spring constant value.
Taking units on both sides
Unit of F = Unit of k x Unit of x
N = Unit of k x m
Unit of k = N/m
SI unit of k (spring constant) = N/m
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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