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2 years ago

Why do we always use water as a reference material when determining the relative density of a substance

Physics

2 answers:

enyata [817]2 years ago

8 0

Answer:

Specific gravity for liquids is nearly always measured with respect to water at its densest

Explanation:

Archy [21]2 years ago

8 0

Answer:

Because of it being able to become a gas, a liquid, and solid, and because its used in lots of chemical experiments. (also because its easier to describe it than any other liquid)

Explanation:

1. Water is used in most Chemicals, therefore its being used as a reference material.

2. W

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2 years ago

A pendulum is raised to a height of 0.3m above its lowest point and released. What is the velocity of the pendulum at its lowest

enyata [817]

Answer:

v = 2,425 m / s

Explanation:

A simple pendulum has anergy stored at the highest point of the path and this energy is conserved throughout the movement.

highest point

Em₀ = U = m g y

lowest point

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

mg y = ½ m v²

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v = √ (2 9.8 0.3)

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3 years ago

Out of aluminum,copper,steel,and glass.Which material do you think will be the best thermal conductor?

Nastasia [14]

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3 years ago

Where do we use friction in day to day life

laila [671]

Answer:

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2 years ago

(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th

Ghella [55]

Answer:

Assuming that the vertical speed of the ball is 14 m/s we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

Explanation:

a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:

$V_{f}^{2} = V_{0}^{2} - 2gh$

Where:

$V_{f}$ : is the final speed = 14 m/s

$V_{0}$ : is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

$V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s$

b) The maximum height is:

$V_{f}^{2} = V_{0}^{2} - 2gh$

$h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m$

c) The time can be found using the following equation:

$V_{f} = V_{0} - gt$

$t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s$

d) The flight time is given by:

$t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s$

I hope it helps you!

3 0

3 years ago