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Gennadij [26K]
2 years ago
15

___________ is the process by which wind removes surface materials.

Physics
2 answers:
Travka [436]2 years ago
4 0
The answer is deflation...have a good day

kherson [118]2 years ago
4 0
<h3><u>Answer;</u></h3>

Deflation

<u>Deflation</u> is the process by which wind removes surface materials.

<h3><u>Explanation</u>;</h3>
  • Erosion refers to a process in which soil, rock, and other surface material are removed from their location and transported to another by natural agents such as water and wind.
  • <em><u>The main mechanism of wind erosion is deflation. Deflation involves erosion by wind of loose material from flat areas of dry uncemented sediments such as those in deserts, dry lake beds, flood plains etc. The removed particles are then transported to another region where they may form sand dunes on a beach or in a desert.</u></em>
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When observing a group of children at a daycare center, Emily made the following observations: Five year old children played in
konstantin123 [22]
This question is not about physics science.

The answer is: option <span>a. Five-year-old children have longer attention spans than three-year-old children.

It is the attention ability what let the older children to stay longer in one location instead of being moving between different activities. The younger children who cannot keep their attention long time in a same activity entertain themselves by changing activities.
</span>
5 0
2 years ago
Read 2 more answers
With what minimum speed must you toss a 110 g ball straight up to just touch the 11-m-high roof of the gymnasium if you release
Sholpan [36]

Answer:

 v = 13.79 m/s

Explanation:

given,

mass of ball = 110 g

height = 11 m

ball is released from = 1.3 m

minimum speed = ?

using conservation of energy

Potential energy is conserved in the form of kinetic energy

P E =KE

m g h= \dfrac{1}{2}mv^2

v^2 = 2 g h

v=\sqrt{2 g h}

v=\sqrt{2\times 9.8 \times (11-1.3)}

v=\sqrt{2\times 9.8 \times 9.7}

v=\sqrt{190.12}

       v = 13.79 m/s

7 0
3 years ago
AYUDAAA PORFAVOR
Sholpan [36]

Queremos crear un diagrama general para calcular el área de un triangulo.

Este será algo como:

  1. Definir variables
  2. Pedirle al usuario que introduzca los valores deseados (de las variables).
  3. Leer los valores deseados y asignarlo a la variable correspondiente.
  4. Realizar la operación para calcular el área.
  5. Mostrar en pantalla el resultado.

Como naturalmente habra algunas variaciones segun el programa que utilicemos, lo voy a escribir de forma bastante general.

Primero definamos nuestras variables:

Por ejemple, en fortran usariamos algo como:

real:: B, H, A

Donde B será la variable que usaremos para la base, H para la altura, y A para el área.

Luego tenemos que escribir en pantalla algo que le diga al usario que debe introducir la base y el area.

Luego el programa debe ser capaz de leer ese input.

con algo de la forma:

B = read*input 1

H = read*input 2

Una vez tenemos definidas las variables, simplemente calculamos el área del triangulo:

A = H*B/2

Finalmente la podemos mostrar en pantalla con algo como:

print(A).

Lo que nos mostraría el valor del área.

Concluyendo, el diagrama en general sería:

  1. Definir variables
  2. Pedirle al usuario que introduzca los valores deseados (de las variables).
  3. Leer los valores deseados y asignarlo a la variable correspondiente.
  4. Realizar la operación para calcular el área.
  5. Mostrar en pantalla el resultado.

Si quieres aprender más, puedes leer:

brainly.com/question/21949109

5 0
2 years ago
The primary reason a light bulb emits light is due to
Mashutka [201]
<span>The primary reason a light bulb emits light is due to the heating of the resistance in the filament of the light bulb. In fact, the power dissipated in a resistor is given by
</span>P=I^2 R
<span>where I is the current and R the resistance. The larger the resistance or the current in the resistor, the larger the power dissipated. Due to this dissipation of power, the temperature of the filament becomes very high, and the resistance becomes incandescent, emitting light.</span>
5 0
3 years ago
Read 2 more answers
An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial
Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2

Mass gets cancelled

-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}

v_e=\sqrt{\frac{2Gm}{R}} = Escape velocity of Earth = 11.2 km/s

v_i = Velocity of projectile = 44.4 km/s

v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

6 0
2 years ago
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