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3 years ago

The light you see around you from the Sun is called UVvisibleX-infrared rays.

1 answer:

6 0

The correct answer for the question that is being presented above is this one: "Ultraviolet rays." The light you see around you from the Sun is called ultraviolet rays. Sunlight is a portion of the electromagnetic radiation given off by the Sun, in particular infrared, visible, and ultraviolet light.

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larisa [96]

Answer:

Aceleración, a = 2 m/s²

Explanation:

Dados los siguientes datos;

Velocidad inicial = 108 km/h

Tiempo = 10 segundos

Velocidad final = 36 km/h

To find the average acceleration;

Conversión:

36 km/h to meters per seconds = 36*1000/3600 = 10 m/s

108 km/h to meters per seconds = 108*1000/3600 = 30 m/s

I. Para encontrar la aceleración, usaríamos la primera ecuación de movimiento;

$V = U + at$

Dónde;

V es la velocidad final.

U es la velocidad inicial.

a es la aceleración.

t es el tiempo medido en segundos.

Sustituyendo en la fórmula, tenemos;

$30 = 10 + a*10$

$30 = 10 + 10a$

$10a = 30 - 10$

$10a = 20$

$Aceleracion = \frac{20}{10}$

Aceleración, a = 2 m/s²

3 0

2 years ago

An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.

jekas [21]

Answer:

$\varphi_1= 796.25 N m^2/C$

$\varphi_2= 0 N m^2/C$

$\varphi_3=686.1 N m^2/C$

Explanation:

From the question we are told that

Electric field of intensity $E= 3.25 kN/C$

Rectangle parameter Width $W=0.350 m$ Length $L=0.700 m$

Angle to the normal $\angle=30.5 \textdegree$

Generally the equation for Electric flux at parallel to the yz plane $\varphi_1$ is mathematically given by

$\varphi_1=EA cos theta$

$\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0$

$\varphi_1= 796.25 N m^2/C$

Generally the equation for Electric flux at parallel to xy plane $\varphi_2$ is mathematically given by

$\varphi_2=EA cos theta$

$\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90$

$\varphi_2= 0 N m^2/C$

Generally the equation for Electric flux at angle 30 to x plane $\varphi_3$ is mathematically given by

$\varphi_3=EA cos theta$

$\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5$

$\varphi_3=686.072219 N m^2/C$

$\varphi_3=686.1 N m^2/C$

7 0

3 years ago

A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the

Alexus [3.1K]

Answer:

the final potential energy of this system is 3U0/10

Explanation:

We are given

charge at left end and another test charge at point p

Potential energy is given by = $\frac{k*Q1*Q2 }{R}$

where k is electrostatics constant = $9 *10^9$

Q1 = first charge , Q2= test charge

R= distance between charges

potential at point p

U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1

now the test charge moves to point R

using Pytahgoreou theorem

R(distance) = $\sqrt{8^2 + 6^2}$ = 10

New Potential energy

U1 = kq1*q2 / 10

substituting kq1q2 = 3U0 from 1

U1 = 3U0/10

So this is the final potential energy of this system.

5 0

2 years ago

A scientist asked a question that was based on an observation. Which is the next step the scientist should take?

ANTONII [103]

In a scientific procedure, the scientist asked a question or several questions based on an observation. After asking such questions, the scientist must form a hypothesis based on the observations made and design an experiment. Hypothesis is an educated guess necessary in a scientific procedure.

5 0

3 years ago

A wind turbine takes in energy from wind with the goal of converting it into electrical energy. Much of the wind energy is also

Shalnov [3]

German physicist Albert Betz (in 1919) demonstrated that the highest efficiency you can achieve with a wind turbine is around 59%

We would have to analyze the design of an specific turbine to determine its efficiency, however it is unlikely to achieve 50% , as todays turbines have an average efficiency in the 20-35%

The answer would be around 25%

6 0

3 years ago

Read 2 more answers