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3 years ago

Which of the following waves are mechanical?

Physics

2 answers:

tensa zangetsu [6.8K]3 years ago

6 0

A. Ocean waves

"please give me brainliess"

spin [16.1K]3 years ago

5 0

Ocean waves would be the correct answer
Your welcome!!!!!!!!!!!!

You might be interested in

Why do you think that some asteroids tumble end over end through space while other asteroids rotate around their axis?

olga2289 [7]

Imagine a lonely asteroid there in space, untouched, not rotating, just still. Then another asteroid passes by and comes into contact with that first asteroid. Upon collision, based on the conservation of elastic momentum, the asteroid that was once still moves; it may even spin if the incoming asteroid hit it at its side. Now asteroids have three angles or rotation (three dimensions): θ (theta - to the x-axis), φ (phi - to the y-axis), and let's say ψ (psi - to the z-axis). So these asteroids wobble through space, spinning like crazy.

I may have gone too in depth. Sorry, lol.
I hope this helped!

7 0

3 years ago

A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the

krek1111 [17]

Answer:

45.6 cm

Explanation:

Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation

$E_g = E_p$

$mgh = 0.5kx^2$

where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.

$4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2$

$47.088x + 227.906 = 1200x^2$

$1200x^2 - 47.088x - 227.906 = 0$

$x = 0.456m$ or 45.6 cm

5 0

4 years ago

Two pans of a balance are 46.3 cm apart. The fulcrum of the balance has been shifted 0.633 away from the center by a dishonest s

soldier1979 [14.2K]

distance of each pan from the center or fulcrum is given as

$r = 23.15 cm$

now if dishonest shopkeeper shifted it by 0.633 cm from center

so distance on each side is given as

$d_1 = 23.15 - 0.633 = 22.52 cm$

$d_2 = 23.15 + 0.633 = 23.78 cm$

now the weight is balance as

$W_1d_1 = W_2d_2$

$W(22.52) = W_2(23.78)$

now we will have

$W_2 = 0.95W$

now we can find the percentage change as

$percentage = \frac{W - W_2}{W} \times 100$

$percentage = 5%$

5 0

4 years ago

A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

Tems11 [23]

Answer:

a) $V_{wf}$ = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

$m_{b}V{b_{i} } + V_{wi} = m_{w} V_{wf} + m_{b}V_{bf}$

where $m_{b},V{b_{i}$ are the mass and the initial velocity of the bullet, $m_{w}$ and $V_{wi}$ are the mass and the initial velocity of the wooden block, and $V_{wf}$ and $V_{bf}$ are the final velocities of the wooden block and the bullet

The wooden block is initial at rest $(V_{wi} = 0)$ this yields

$m_{b}V{b_{i} } = m_{w} V_{wf} + m_{b}V_{bf}$

By solving for $V_{wf}$ adn substitute the givens

$V_{wf}$ = $\frac{m_{b}V_{bi} - m_{b} V_{bf} }{m_{W} }$

= $\frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}$

$V_{wf}$ = 4.67m/s

b) The center of mass speed is defined as

$V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}$

substituting:

$V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)$

V = 8.29 m/s

7 0

3 years ago

Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45

victus00 [196]

<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>

4 0

3 years ago