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tatiyna
2 years ago
5

8. When wind dies down or stops blowing happens.

Physics
2 answers:
timama [110]2 years ago
5 0

Answer:

Deposition happens, if the wind stops blowing.

Deposition is the dropping of sediment by wind, water, ice, or gravity.

yan [13]2 years ago
4 0

Answer:

Deposition

Explanation:

  • Deposition occurs when water slows or ceases moving, the wind dies or stops blowing, or glaciers melt. The deposited material can also be used to construct new landforms. Waves, for example, can dump sediment in places offshore, where it might accumulate to form sand dunes.

When the wind calms down or vegetation stops or slows the breeze, the sediment particles begin to fall. Water is another factor that may erode, move, or deposit sediment. Flowing water is a key erosive agent. Water transports dirt and rock fragments as it moves. Warm, wet air will not travel if wind systems are not present. Water will still evaporate, but it will not move, therefore everywhere else than a major body of water will dry up. Lakes may be fine since evaporating water will flow back into them, and the sea will be fine, but everywhere else will become extremely dry very rapidly. Wind is constantly blowing somewhere on the world at any given time. Winds are usually quiet near the middle of a high pressure system. Wind is the passage of air from a high pressure location to a low pressure area.... So essentially air is always moving. Weathering and erosion are caused by wind. Weathering is caused by wind blowing debris against cliffs and huge rocks. This wears down the rock, reducing it to sand and dust. Sand and dust are also eroded by wind. 2. Rocks are tough and durable, but they don't last forever. Weathering and erosion are processes that occur as a result of forces such as wind and water breaking down rocks. Weathering is the process through which rocks deteriorate. Weathering is caused by a variety of factors, including climate change.

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A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

3 0
3 years ago
a person does 100 joules of work in pulling back the string of a bow. what will be the initial speed of a 0.5 kg arrow when it i
eduard

If all the energy she put into bending the bow is completely
transmitted to the arrow, then the arrow has the 100 joules
of kinetic energy when it leaves the bow.

                 Kinetic energy = (1/2) (mass) (speed)²

                         100 J        = (1/2) (0.5 kg) (speed²)

Divide each side by 0.25 kg:    100 J / 0.25 kg = speed²

 [ joule ] = [ newton-meter ] = kg-m²/sec²

                                                100 kg-m²/sec² / 0.25 kg = speed²

                                                 400 m²/sec² = speed²

Take the square root of each side:     speed = √400 m/s

                                                                                20 m/s

                                                                    (about 44.7 mph)                                           

3 0
2 years ago
What is the force due to gravity of a 38 kg student?
alukav5142 [94]

Answer:

F_g = 372.78 N

Explanation:

Formula for force of gravity is;

F_g = mg

Where;

m is mass

g is acceleration due to gravity

We are given;

Mass = 38 kg

Acceleration due to gravity has a constant value of 9.81 m/s²

Thus;

F_g = 38 × 9.81

F_g = 372.78 N

6 0
3 years ago
While walking past a construction site, a person notices a pipe sticking out of a second floor window with water pouring out. As
tia_tia [17]

Answer:

Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.

P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0

Explanation:

P1 = P2 = atmospheric pressure so, P1 - P2 = 0

h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²

From the continuity equation for fluids

A1v1 = A2v2

v2 = A1v1/A2

Substituting into the equation above

(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho

Making A2² the subject of the formula,

A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}

The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.

Thank you for reading.

4 0
2 years ago
Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es
vfiekz [6]

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

3 0
3 years ago
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