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Tom [10]
3 years ago
5

What instrument will we use to find liquid volume

Physics
1 answer:
Elanso [62]3 years ago
3 0

Answer:

Volumetric cylinders (would appreciate brainliest

Explanation:

Volumetric cylinders and volumetric flasks are used to measure volume of liquids contained in them. They are calibrated for volume included in them - this is indicated by the marking "IN". The liquid has accurate volume when it reaches the corresponding marking on the scale.

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What part of an atom is responsible for producing magnetic fields
Nezavi [6.7K]
Electrons is the part of the atom that creats a magnetic feild

6 0
3 years ago
What happens when two aqueous solutions are combined in a precipitation reaction and no precipitate is formed?   A. The solution
inysia [295]

There's no reaction for me.

5 0
3 years ago
edward travels 150 kilometers due west and then 200 kilometers in a direction 60 degrees north of west.what is his displacement
Naily [24]

The displacement of Edward in the westerly direction is determined as 338.32 km.

<h3>What is displacement of Edward?</h3>

The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.

The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰

The displacement is the side of the triangle facing 150⁰ = R

R² = a² + b² - 2abcosR

R² = 150² + 200²  - (2x 150 x 200)xcos(150)

R² = 62,500 - (-51,961.52)

R² = 114,461.52

R = 338.32 km

Learn more about displacement here: brainly.com/question/321442

#SPJ1

3 0
1 year ago
if the coefficient of linear expansion of a metal is 2.05× 10^-6 k^-1 what will be its new length if 50cm metal went through a t
boyakko [2]

Answer:

L = L0 (1 + c T)   where c is the coefficient and T the change in temperature

L = 50 ( 1 + 2.05E-6 * 50) = 50.0051 cm

7 0
2 years ago
If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re
-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = \frac{1}{2}kx^2   ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = \frac{1}{2}k\times0.021^2

or

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = \frac{1}{2}\times13605.44\times (0.062)^2

or

U = 26.149 J

Also,

Work done = change in energy

or

W = 26.149 - 3.0 = 23.149 J

4 0
3 years ago
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