All categories

3 years ago

What is the most common source of energy for surface waves

Physics

1 answer:

Elanso [62]3 years ago

5 0

Answer:

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

Explanation:

You might be interested in

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

Goodmorning I can’t seem to remember this can someone help me

Bumek [7]

Explanation:

3 cause the triangle method of addition are connected to tips of one vector

6 0

3 years ago

Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work

LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

$\frac{\Delta x}{\Delta t}$

Analysing the units, meters (displacement) and seconds (time) are basic units:

$\frac{m}{s}$

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

$F = ma$

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

$a=\frac{\Delta v}{\Delta t}$

Analysing the units:

$\frac{\frac{m}{s}}{s}=\frac{m}{s^2}$

Therefore, the unit of force is:

$kg\frac{m}{s^2}$

Pressure:

Pressure is given by the equation:

$P=\frac{F}{S}$ where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

$[l^2]=m^2$ , the unit of area

Dividing the aforementioned unit of force by the unit of area:

$\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}$ , the unit of pressure

Work:

Work is given by the equation:

$W=\vec{F}\cdot \vec{x}$ , (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

$[W]=\frac{kgm}{s^2}m=\frac{kgm^2}{s^2}$ , the unit of work.

3 0

3 years ago

An automobile with an initial speed of 5.13 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed of the car a

Digiron [165]

Answer:

Final velocity v=19.83 m/sec

Explanation:

We have given initial velocity u =5.13 m/sexc

Acceleration of automobile $a=3m/sec^2$

Time t =4.9 sec

We have to find the final velocity v

According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time

So $v=5.13+3\times 4.9=19.83m/sec$

So the final velocity is 19.83 m/sec

7 0

3 years ago

A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol

qaws [65]

Answer:

Part a)

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

$d = \frac{1}{2}gt^2$

Part c)

$v_f = v_o - gt$

Part d)

$d = \frac{1}{2}gt^2$

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

$v_y = -gt$

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

$d = \frac{1}{2}gt^2$

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

$v_f = v_i + at$

$v_f = v_o - gt$

Part d)

distance from helicopter

$d = \frac{1}{2}gt^2$

8 0

3 years ago

Read 2 more answers