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3 years ago

12

The process by which beach sediment moves down the beach with the current is called

1 answer:

irinina [24]3 years ago

6 0

Answer:

Longshore drift

Explanation:

As waves repeatedly hit the beach, some of the beach sediment moves down the beach with the current, in a process called Longshore drift. A spit is a beach that projects like a finger out into the water. A sand dune is a deposit of wind-blown sand.

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HELP PLSSSSSSS ASPA 20 pointsssss

FrozenT [24]

Answer:

Hey guess what there's no question genius

Explanation:

8 0

3 years ago

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A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

4 years ago

How are Neptune and Uranus allke?

AnnZ [28]

Answer:

The methane gives Neptune the same blue color as Uranus.

Explanation:

4 0

3 years ago

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A uniform disk of mass 21 kg, thickness 0.5 m, and radius 0.6 m is located at the origin, oriented with its axis along the y axi

goldenfox [79]

Answer:

$29.69 kgm^2/s$

Explanation:

So suppose the axis of rotation is perpendicular to the surface of the disk, then the moment of inertia can be calculated as the following:

$I = mr^2/2 = 21 * 0.6^2/2 = 3.78 kgm^2$

We can convert the rotation speed in term of 0.8 seconds per revolution to the angular velocity knowing that each revolution is 2π

$\omega = 2\pi / 0.8 = 7.85 rad/s$

Then the rotational angular momentum of the disk is:

$\omega I = 7.85 * 3.78 = 29.69 kgm^2/s$

In case the axis of rotation is parallel with the surface, the moment of inertia would have a formula of:

$I = m(3r^2 + h^2)/12 = 21*(3*0.6^2 + 0.5^2)/12 = 2.3275 kg m^2$

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3 years ago

The diagram below shows the stars that are nearest to our solar system.

abruzzese [7]

Answer:

no u tried of the same dam

thing

Explanation:

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4 years ago