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3 years ago

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What would be a good indicator of the solubility of a substance in a solute? A. whether the substance is a solid a liquid or a g

as B. the amount of substance needed to form a saturated solution C. the density of the substance D. how well the substance dissolved in a different solvent

1 answer:

Fed [463]3 years ago

6 0

The density of the substance

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A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

Give 10 examples of units you might use or see in any given day

Ierofanga [76]

Cups
teaspoon
tablespoon
liters
milliliters
gallons
pints
tons
inches

3 0

3 years ago

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second

shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is 27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC = 25cos 30

OE = 25sin30

OE = CD

Therefore BC = 30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in $\Delta BCO$

$OB^2 = BC^2 + OC^2$

PUTTING ALL VALUE IN ABOVE EQUATION

$d^2 = 930-25sin30)^2 + (25cos30)^2$

$d^2 = 775$

d = 27.83 Km

distance between both oasis ( 1 and 2) is 27.83 Km

3 0

3 years ago

Copernicus and other astronomers before him thought that celestial bodies followed a _____ orbital path.

murzikaleks [220]

The correct answer is circular. Copernicus and other astronomers before him thought that celestial bodies followed a circular orbital path. Copernicus was a Polish astronomer that concluded that the sun is at rest near the center of the universe and the earth is revolving around it annually. This theory is called heliocentric.

6 0

2 years ago

Read 2 more answers

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = $m_1=6.50 kg- 1.55 kg=4.95 kg$

Velocity achieved by squid = $v_1=2.60 m/s$

Momentum gained by squid = $P=m_1v_1$

Mass of water = $m_2=1.55 kg$

Velocity by which water was released by squid = $v_2$

Momentum gained by water but in opposite direction = $P'=m_2v_2$

P = P'

$m_1v_1=m_2v_2$

$v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s$

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E = $\frac{1}{2}m_1v_1^{2}$

Kinetic energy of water = K.E' = $\frac{1}{2}m_2v_2^{2}$

Total kinetic energy created by this maneuver:

$K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}$

$=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules$

4 0

3 years ago