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Blababa [14]
2 years ago
12

1. Is there a relationship between the volume of water displaced and the total volume of the block that has anything to do with

density? If so, what is it?
Physics
1 answer:
strojnjashka [21]2 years ago
8 0

Answer:

The volume of the block is equal to the volume of water displaced by the block.

Explanation:

Volume refers to the amount of space occupied by a given object (in this case the block). When an object such as the block is immersed in water, it displaces its own volume of water. This volume of water displaced is equal to the volume of the block. Hence we can write;

Final Volume of water - Initial Volume of water= Water Displaced = Volume of the block

Recall that the density of a body is given by;

Density= mass/volume

If we obtain the volume of the block by measuring the volume of water displaced by the block, then we weigh the block using a weighing balance, we can obtain the density of the block easily from the relationship shown above.

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Atoms contain empty space true or false​
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Magnet A has twice the magnetic field strength of magnet B and pulls on magnet B with a force of 100 N. The amount of force that
son4ous [18]

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F\propto B

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\begin{gathered} F_B\propto B_B \\ F_B\propto\frac{B_A}{2} \\ F_B=\frac{F_A}{2} \\ F_B=\frac{100}{2} \\ F_B=50\text{ N} \end{gathered}

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3 0
11 months ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

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given data

voltage = 3 kV

magnetic field = 0.66 T

solution

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PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

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v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

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a = \frac{3.82\times 10^{18}}{9.81}  

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and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
Wtz the answer to this question? I need to get it right
Rama09 [41]

Answer: Next time you create a question, add an image or PDF. Because I do not know the question. So, may you please create a new question?

4 0
2 years ago
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