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2 years ago

What is the most common source of energy for surface waves

Physics

1 answer:

Elanso [62]2 years ago

5 0

Answer:

The most common source of energy for surface waves is wind. earthquakes. tides. volcanoes.

Explanation:

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If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth

fgiga [73]

<u>Answer:</u> The elevation in boiling point is 1.024°C.

<u>Explanation:</u>

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=ik_b\times m$

where,

i = Van't Hoff factor = 2 (for NaCl)

$\Delta T_b$ = change in boiling point = ?

$k_b$ = boiling point constant = $0.512^oC/m$

m = molality = 1.0 m

Putting values in above equation, we get:

$\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC$

Hence, the elevation in boiling point is 1.024°C.

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3 years ago

Rosný bod závisí od ?

WITCHER [35]

Answer:

what did u say and what language are you speaking in

8 0

2 years ago

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi

Marrrta [24]

Answer:

h = 2.64 meters

Explanation:

It is given that,

Mass of one ball, $m_1=3\ kg$

Speed of the first ball, $v_1=20\ m/s$ (upward)

Mass of the other ball, $m_2=2\ kg$

Speed of the other ball, $v_2=-12\ m/s$ (downward)

We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :

$V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$

$V=\dfrac{3\times 20+2\times (-12)}{3+2}$

V = 7.2 m/s

Let h is the height reached by the combined balls of putty rise above the collision point. Using the conservation of energy as :

$mgh=\dfrac{1}{2}mV^2$

$h=\dfrac{V^2}{2g}$

$h=\dfrac{7.2^2}{2\times 9.8}$

h = 2.64 meters

So, the height reached by the combined mass is 2.64 meters. Hence, this is the required solution.

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2 years ago

Consider your knowledge and experience about caring for children. What have you observed about the needs, behaviors, and abiliti

Advocard [28]

They all have a special way in their age how they want to get taught things but there abilities make them special

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2 years ago

A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th

ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

-Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

a is the acceleration of the car

a = $\frac{frictional force}{mass}$ = $\frac{10500}{1500}$ = -7m/s²

Now using;

V² = U² + 2as

V is the final velocity

U is the initial velocity

a is the acceleration

s is the distance moved

0² = 25² + 2 x 7 x s

0 = 625 - 14s

-625 = -14s

s = 44.64m

learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

7 0

3 years ago