Alejandro camina 250 metros en linea recta durante 5 minutos y regresa a su mismo camino a su punto de partida en 7 minutos. ¿cual es la distancia total recorrida? ¿cual es la velocidad en m/s? ¿cuanto vale su desplazamiento total?

Answers

Answer 1

Alejandro walks 250 meters in a straight line for 5 minutes and returns to his same path to his starting point in 7 minutes.
Going to the point, he walked 50 meters every minute and going back, he walked 35.71428571428571

Related Questions

How many valence electrons does the hydrogen and oxygen have in the following picture?

Answers

Answer:

can u make the picture less blurry pls

Explanation:

A coconut at rest suddenly explodes into two pieces. no net external force is applied to it.One piece with mass 2m lands a distance d to the right of the starting point. The second piece has mass m. We can ignore air resistance. where dow stew swcond piece land?

Answers

Answer:

To the Left

Explanation:

why are antennas needed for radio, television, and cell phone transmission>

A. without a long antenna, the signal will not travel very far
B. the antenna allows you to direct your signal toward the intended target
C. the sound waves require a metal surface to bounce off
D. EM waves are created by oscillating electrons within the antenna
40 points + brainliest if correct please help

Answers

Answer:

A.without a long antenna, 're signal will not travel far

The distance from the center of the lens to its principal focus is called the focal length.
O A. True
O B. False​

Answers

Answer:

true

Explanation:

Answer:

Explanation:

A. True

which subatomic particle is transferred through circuits

Answers

Answer:A third type of subatomic particle, electrons, move around the nucleus. The electrons have a negative electrical charge. An atom usually contains an equal number of positively charged protons and negatively charged electrons.

Explanation:

If 30.45 grams of water is to be heated up 3.3 degrees to make baby
formula, how much heat must be added? The specific heat of water is
4.18 J/g °C.

A. 420 J
B. 281 J
C. 1604 J
D. 410 J

Answers

Answer:

A. 420 J

Explanation:

Given the following data;

Mass = 30.45 g

Specific heat capacity = 4.18 J/g °C.

Temperature = 3.3°C

To find the quantity of heat;

Heat capacity is given by the formula;

[tex] Q = mct [/tex]

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

t represents the temperature.

Substituting into the equation, we have;

[tex] Q = 30.45 * 4.18 * 3.3 [/tex]

Q = 420.03 ≈ 420 Joules

Which material is a conductor?
A. chalk
B. lead
C. leather
D. paper
E. rubber

Answers

B. Lead is a conductor....

Un móvil posee un movimiento uniformemente acelerado, con una velocidad inicial de 20 m/s y aceleración 1,5 m/s. ¿Qué velocidad tendrá cuando hayan transcurrido 2 minutos? ¿Qué espacio habrá recorrido durante ese tiempo?

Answers

Answer:

La velocidad que tendrá el móvil cuando hayan transcurrido 2 minutos es 200 m/s.

El espacio que habrá recorrido el móvil durante 2 minutos es 13200 metros.

Explanation:

El movimiento rectilíneo uniformemente variado o MRUV es un movimiento que ocurre sobre una línea recta con aceleración constante. En otras palabras, un cuerpo realiza un  movimiento rectilíneo uniformemente variado cuando su trayectoria es una línea recta y su aceleración es constante y distinta de 0. Esto implica que la velocidad aumenta o disminuye su módulo de manera uniforme.

La velocidad que tendrá un móvil luego de un tiempo t será:

v= v0 + a*t

donde v0 es la velocidad inicial y a es la aceleración.

En este caso v0= 20 m/s, a=1,5 m/s² y t=2 minutos= 120 segundos. Reemplazando y resolviendo se obtiene:

v= 20 m/s + 1.5 m/s²* 120 segundos

v= 200 m/s

La velocidad que tendrá el móvil cuando hayan transcurrido 2 minutos es 200 m/s.

La expresión de la  posición en función del tiempo para el movimiento uniformemente variado es:

x= x0 + v0*t + [tex]\frac{1}{2}[/tex] *a*t²

donde x0 es la posición inicial, v0 es la velocidad inicial y a es la aceleración.

En este caso x0= 0 m, v0= 20 m/s, a=1,5 m/s² y t=2 minutos= 120 segundos. Reemplazando y resolviendo se obtiene:

x= 20 m/s* 120 s +  [tex]\frac{1}{2}[/tex] * 1.5 m/s² *(120 s)²

x= 2400 m + 10800 m

x= 13200 m

El espacio que habrá recorrido el móvil durante 2 minutos es 13200 metros.

???????????!??!!!!!!!!​

Answers

The answer would be B.

What type of lens is used in the objective lens (front lens) of binoculars and why?
(Concave or convex)

Answers

Answer:

If you want to see something in the distance, you can use two convex lenses, placed one in front of the other. The first lens catches light rays from the distant object and makes a focused image a short distance behind the lens. This lens is called the objective, because it's nearest to the object you're looking at.

Explain why all living have genes and DNA but not all living things look the same. (Discuss, paragraph answers)

Answers

Answer:

All living organisms store genetic information using the same molecules DNA and RNA. Genes are maintained over an organism's evolution, however, genes can also be exchanged or "stolen" from other organisms.

A man is driving his 1100 kg car at 36 km/h on a straight freeway. After accelerating for 30 seconds, the car
has a speed of 108 km/h. How much work did the engine do during the 30 seconds?

Answers

Answer:

W = 439998 J = 439.99 KJ

Explanation:

First, we will calculate the acceleration of the car by using the first equation of motion:

[tex]v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}[/tex]

where,

a = acceleration = ?

vf = final speed = [tex]108(\frac{km}{h})(\frac{1000\ m}{1\ km})(\frac{1\ h}{3600\ s})[/tex] = 30 m/s

vi = initial speed = [tex]36(\frac{km}{h})(\frac{1000\ m}{1\ km})(\frac{1\ h}{3600\ s})[/tex] = 10 m/s

t = time = 30 s

Therefore,

[tex]a = \frac{30\ m/s - 10\ m/s}{30\ s}[/tex]

a = 0.67 m/s²

Now, we will calculate the force applied by the engine:

F = ma

where,

F = force = ?

m = mass = 1100 kg

Therefore,

F = (1100 kg)(0.67 m/s²)

F = 733.3 N

Now, we will calculate the distance covered by the car by using the second equation of motion:

[tex]s = v_it+\frac{1}{2}at^2\\\\s = (10\ m/s)(30\ s)+\frac{1}{2} (0.67\ m/s^2)(30\ s)^2[/tex]

s = 600 m

Now, the work done (W) by engine can be calculated as follows:

W = Fs

W = (733.3 N)(600 m)

W = 439998 J = 439.99 KJ

The gravitational force is 2336 N for an object that is 4.15 x 10^6 m above the surface of the Earth? The radius of the Earth is 6.378 x 10^6 m. (Earth's mass is 5.97 x 10^24 kg) What is the mass of the object above earth?

Answers

Answer:

34kg

Explanation:

A planet orbits a sun in a clockwise elliptical orbit as shown in the diagram below
Bubble in the letter representing the position or choose e if it is constant

At which position does the planet have the greatest gravitational energy?

Answers

Answer:

Greatest gravitational energy is at "C".

The planet has to do work "against" the field to get to "C".

Also, if m v R (angular momentum) is constant then as R increases v must decrease for this term to be constant and KE = 1/2 M v^2  must decrease also to get to point C.

In the diagram, R1 = 40.0 ohm, R2 = 25.4 ohm, and R3 = 70.8 ohm. What is the equivalent resistance of the group?​

Answers

Given :

[tex]R_1 = 40 ohms[/tex]

[tex]R_2 = 25.4 ohms[/tex]

[tex]R_3 = 70.8 ohms[/tex]

Solution :

The resistors [tex]R_1 and R_2 [/tex]are in series connection, so the equivalent resistance of [tex]R_1 and R_2[/tex] will be their sum.

[tex] \boxed{ \mathrm{R_t = R_1 + R_2}}[/tex]

[tex]R_t= 40 + 25.4[/tex]

[tex]R_t= 65.4 \: \: ohms[/tex]

Now, the equivalent resistance of [tex](R_1 and R_2)[/tex] and [tex]R_3 [/tex] is the total resistance of the circuit, and since they are in parallel connection, Total resistance :

[tex] \mathrm{ \dfrac{1}{ R_{eq}}} = \dfrac{1}{65.4} + \dfrac{1}{70.8} [/tex]

[tex] \dfrac{1}{R_{eq}} = \dfrac{65.4 + 70.8}{4630.32} [/tex]

[tex] \frac{1}{R_{eq}} = \dfrac{136.2}{4630.32} [/tex]

[tex]R_{eq} = \dfrac{4630.32}{136.2} [/tex]

[tex]R_{eq} = 33.996 \: \: ohms[/tex]

[tex] \boxed{ \mathrm{R_{eq} = 34 \: ohms}} \: (approx)[/tex]

_____________________________

[tex]\mathrm{ \#TeeNForeveR}[/tex]

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