Answer:
b)7.89 g
c) 7.91 g
d) 7.97 g
Explanation:
The accepted value for the measurement is 7.92 g, while the precision of the instrument is 0.05 g. Therefore, the interval of the accepted values will be the central value plus and minus the precision, so:
- Minimum accepted value: 7.92 g - 0.05 g = 7.87 g
- Maximum accepted value: 7.92 g + 0.05 g = 7.97 g
So, the accepted interval is
7.87 g - 7.97 g
Therefore, all the values that lie within this interval are accepted. So the accepted values are:
b)7.89 g
c) 7.91 g
d) 7.97 g
I.i = j.j = 1
i.j = j.i = 0
(-2i - 5j)(i -4j)
= (-2i).i + (-2i).(-4j) + (-5j).i + (-5j).(-4j)
= -2 + 0 + 0 + 20 = 18
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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