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3 years ago

7

Whats the similarities of rogue waves and tsunamis

2 answers:

Trava [24]3 years ago

7 0

<span>Rogue wave is caused in open water not caused by earthquake. Tsunami is caused by an earthquake, bottom of the ocean/water.</span>

klio [65]3 years ago

7 0

Answer:

they are dangrous to boats

Explanation:

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Explain the difference between cool air and warm air

Naddik [55]

Answer: ok

Explanation:

The molecules in hot air are moving faster than the molecules in cold air. Because of this, the molecules in hot air tend to be further apart on average, giving hot air a lower density. That means, for the same volume of air, hot air has fewer molecules and so it weighs less.

3 0

3 years ago

According to the graph, what is the motorcycle’s Average velocity ?

borishaifa [10]

According to the position vs time graph, the <em>average</em> <em>velocity</em> of the motorcycle is the change in position divided by the change in time. Also, note that the slope is linear and positive throughout the 5 hours, it doesn't change direction.

Therefore, we have
Avg velocity = change in direction/change in time
Avg velocity = (150km - 30km)/(5h - 0h)
Avg velocity = 24km/hr south.

3 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

3 years ago

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

$v_{1x} = 406 mph\\v_{1y} = 0$

While the components of the velocity of the wind are

$v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph$

So the components of the resultant velocity of the jet are

$v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph$

And the new speed is the magnitude of the resultant velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph$

6 0

3 years ago

A draft is made of a plastic block with a density of 650kg/m^3, and its dimensions are 2.0mx3.0mx5.0m ,what is the raft's appare

mixas84 [53]

Weight in water = mass of block - mass of volume of displaced water

8 0

3 years ago