Enunciado del ejercicio n° 1
Se lanza un cuerpo verticalmente hacia abajo con una velocidad inicial de 7 m/s.

a) ¿Cuál será su velocidad luego de haber descendido 3 s?

b) ¿Qué distancia habrá descendido en esos 3 s?

c) ¿Cuál será su velocidad después de haber descendido 14 m?

d) Si el cuerpo se lanzó desde una altura de 200 m, ¿en cuánto tiempo alcanzará el suelo?

e) ¿Con qué velocidad lo hará?

Usar g = 10 m/s²

Desarrollo
Datos:
v0 = 7 m/s

t = 3 s

y = 200 m

h = 14 m

Fórmulas:
(1) vf = v0 + g·t

(2) y = v0·t + ½·g·t²

(3) vf² - v0² = 2·g·h


Enunciado del ejercicio n° 2
Se lanza un cuerpo verticalmente hacia arriba con una velocidad inicial de 100 m/s, luego de 4 s de efectuado el lanzamiento su velocidad es de 60 m/s.

a) ¿Cuál es la altura máxima alcanzada?

b) ¿En qué tiempo recorre el móvil esa distancia?

c) ¿Cuánto tarda en volver al punto de partida desde que se lo lanzo?

d) ¿Cuánto tarda en alcanzar alturas de 300 m y 600 m?

Usar g = 10 m/s²

Desarrollo
Datos:
v0 = 100 m/s

vf = 60 m/s

t = 4 s

y1 = 300 m

y2 = 600 m

Fórmulas:
(1) vf = v0 + g·t

(2) y = v0·t + ½·g·t²

(3) vf² - v0² = 2·g·h

Enunciado del ejercicio n° 3
Un observador situado a 40 m de altura ve pasar un cuerpo hacia arriba con una cierta velocidad y al cabo de 10 s lo ve pasar hacia abajo, con una velocidad igual en módulo pero de distinto sentido.

a) ¿Cuál fue la velocidad inicial del móvil?

b) ¿Cuál fue la altura máxima alcanzada?

Usar g = 10 m/s²

Desarrollo
Datos:
t = 10 s

y = 40 m

Fórmulas:
(1) vf = v0 + g·t

(2) y = y0 + v0·t + ½·g·t²

(3) vf² - v0² = 2·g·h



Enunciado del ejercicio n° 4
Desde un 5° piso de un edificio se arroja una piedra verticalmente hacia arriba con una velocidad de 90 km/h, ¿cuánto tardará en llegar a la altura máxima?

Usar g = 10 m/s²

Desarrollo
Datos:
v0 = 90 km/h

v0 = 25 m/s

Fórmulas:
(1) vf = v0 + g·t

(2) y = v0·t + ½·g·t²

(3) vf² - v0² = 2·g·h


Enunciado del ejercicio n° 5
Un auto choca a 60 km/h contra una pared sólida, ¿desde qué altura habría que dejarlo caer para producir el mismo efecto?

Usar g = 10 m/s²

Desarrollo
Datos:
vf = 60 km/h

vf = 16,67 m/s

v0 = 0 m/s

Fórmulas:
(1) vf = v0 + g·t

(2) y = v0·t + ½·g·t²

(3) vf² - v0² = 2·g·h

Enunciado del ejercicio n° 6
Se lanza una pelota hacia arriba y se recoge a los 2 s, calcular:

a) ¿Con qué velocidad fue lanzada?

b) ¿Qué altura alcanzó?

Usar g = 10 m/s²

Desarrollo
Datos:
t = 2 s

Fórmulas:
(1) vf = v0 + g·t

(2) y = v0·t + ½·g·t²

(3) vf² - v0² = 2·g·h


Enunciado del ejercicio n° 7
Se lanza una pelota de tenis hacia abajo desde una torre con una velocidad de 5 m/s.

a) ¿Qué velocidad tendrá la pelota al cabo de 7 s?

b) ¿Qué espacio habrá recorrido en ese tiempo?

Usar g = 10 m/s²

Desarrollo
Datos:
v0 = 5 m/s

t = 7 s

Fórmulas:
(1) vf = v0 + g·t

(2) y = v0·t + ½·g·t²

(3) vf² - v0² = 2·g·h


Enunciado del ejercicio n° 8
Se lanza un cuerpo verticalmente hacia arriba con una velocidad de 60 km/h, se desea saber la altura máxima alcanzada, la velocidad que posee al cabo de 4 s y 30 s, la altura alcanzada a los 8 s, el tiempo total que se encuentra en el aire.

Desarrollo
Datos:
v0 = 60 km/h = (60 km/h)·(1.000 m/km)·(1 h/3.600 s) = 16,67 m/s

t1 = 4 s

t2 = 30 s

t3 = 8 s

Usar g = 10 m/s²

Fórmulas:
(1) vf = v0 + g·t

(2) y = v0·t + ½·g·t²

(3) vf² - v0² = 2·g·h


Enunciado del ejercicio n° 9
Se dispara verticalmente hacia arriba un objeto desde una altura de 60 m y se observa que emplea 10 s en llegar al suelo. ¿Con que velocidad se lanzo el objeto?

Desarrollo
Datos:
h0 = 60 m

t = 10 s

g = 9,81 m/s²

Fórmulas:
Δy = v0·t + ½·g·t²


Enunciado del ejercicio n° 10
Se lanza verticalmente hacia abajo una piedra de la parte alta de un edificio de 14 pisos, llega al suelo en 1,5 s, tomando en cuenta que cada piso mide 2,6 m de altura. Calcular la velocidad inicial de la piedra y la velocidad al llegar al piso.

Desarrollo
Datos:
Número de pisos = 14

Altura de cada piso = 2,6 m

t = 1,5 s

g = 9,81 m/s²

Fórmulas:
1) Δh = v0·t + ½·g·t²

2) vf = v0 + g·t


*xfv se que es mucho pero e visto videos pero no me sale muy bien los resultados con mis compañeros. xfv alguien que me ayude

Answer:

u have to purchase it via online e-commerce platforms

Answer:

what your crazy anong pong common paper

Explanation:

A lightning rod (US, AUS) If lightning hits the structure:-

It will preferentially strike the rod and be conducted to ground through a wire.

Instead of passing through the structure, where it could start a fire or cause electrocution.

The parts of a lightning protection system are air terminals (lightning rods or strike) and all of the connectors and supports to complete the system.

Answer:

Hope this is what you were looking for!

Explanation:

1) So, if we are talking about balancing a straight stick (beam), he could put a finger from each hand on either side of it and move them together. As a result of friction his fingers will meet at the balance point.

2) Method of trial and error.

Answer:

False:  Wave lengths on the electromagnetic spectrum are "electromagnetic",

They are transverse waves,

Sound waves are "longitudinal" waves and depend on the motion of the medium of transmission for propagation.

Answer:

it begins to move to the right! :)

Answer:

the force of attraction between all masses in the universe

Explanation:

Have a nice day :)

Answer:

more acidic (less base)

hope this helps! :)

Answer:

solve it with the formula 1−0.4=0.6, 0.6Vρg=Vρbg where ρb 

Explanation:

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=⇒ρb=0.6ρ.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=⇒ρb=0.6ρ.In the second case: 1.5fVρg=Vρbg+Vρba=1.5Vρbg⇒fρ=ρb⇒f=0.6.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=⇒ρb=0.6ρ.In the second case: 1.5fVρg=Vρbg+Vρba=1.5Vρbg⇒fρ=ρb⇒f=0.6.Thus, the fraction of immersed volume remains the same.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=⇒ρb=0.6ρ.In the second case: 1.5fVρg=Vρbg+Vρba=1.5Vρbg⇒fρ=ρb⇒f=0.6.Thus, the fraction of immersed volume remains the same.Body will float with 40% of the volume above water surface.

Answer:

diffraction phenomenon

Explanation:

When you place your finger in front of a flash, you have a diffraction phenomenon whereby the light fills a part of the space with the first maximum diffraction.

In this case the phenomenon of geometric optics cannot explain the process of bending the beam around an object.

Explanation:

the PE = Weight x g x Height above ground where g is the gravitational constant (9.8m/s^2)

so tge PE = 100 x 9.8 x 25 = 24500 J

Answer:

Consider an isolated system of 2 bodies A & B. An isolated system means where there is no external force acting.Now let F be the force acting on B by A & F be the force acting on A by B.Now rate of change of momentum of A =dpA/dt and rate of change of momentum of B =d pB/dt

thus FAB=d pB/dt (i)......

FBA=dpA/dt (ii)........

Adding 1 and 2 we get FAB+FBA=d pB/dt+dpA/dt =d(pB+pA)/dt

But if no force is applied momentum will also be 0 because no velocity will be there so rate of change of momentum will also be 0

thus d(pA+pB)/dt=0

therefore FAB+FBA=0

or,FAB=-FBA

(3RD LAW OF MOTION)

Answer:

Negative acceleration occurs when the acceleration vector points to the left.

1. Object slowing down in the positive direction.

2. Object speeding up in the  negative direction.

Following six statements:

1. T

2. F

3. T

4. T

5. F

6. T

Check direction of acceleration vector.

Answer:

the distance traveled by the sports car is 90 m

Explanation:

Given;

mass of the sports car, m = 500 kg

initial velocity of the sports car, u = 0

final velocity of the sport car, v = 30 m/s

time of motion of the car, t = 6 s

The distance traveled by the sports car is calculated as;

[tex]s = (\frac{u+v}{2} )t\\\\s = (\frac{0+30}{2} ) \times 6\\\\s = 15 \times 6\\\\s = 90 \ m[/tex]

Therefore, the distance traveled by the sports car is 90 m

Answer:

W = 170.62 N

Explanation:

The given question is, "calculate the weight of a person traveling to the moon whose mass is 105 kg".

We need to find the weight of a person on the Moon.

The weight of an object is given by :

W = mg

Where

g is acceleration due to gravity on Moon, g = 1.625 m/s2

W = 105 kg × 1.625 m/s²

W = 170.62 N

Hence, the weight of the person is 170.62 N.

Answer:

1.47 J/kg

120 m

Explanation:

The maximum height it can jump in a single leap 15 cm, = 0.15 m

to get the kinetic energy per kg of mass, we need to find the takeoff speed. The take off speed can be calculated by using the formula

v = [tex]\sqrt{2gh}[/tex]

where h = 0.15

v = [tex]\sqrt{2 * 9.81 * 0.15}[/tex]

v = [tex]\sqrt{2.94}[/tex]

v = 1.715 m/s

energy per kg of mass =

1/2 * (1.715) = 1.47 J/Kg

TJe height a human can jump when compared to the flea is

2 m/2.5 mm * 15 cm =

2/0.0025 *0.15 =

800 * 0.15 = 120 m

Answer:

5

Explanation:

20 x  5 = 100

Answer:

Se requiere corriente para que brille la luz.

Explicación:

Una batería de 1,5 voltios está conectada a una luz LED roja, el LED se enciende siempre que esté conectado a la batería. A partir de esta situación se afirma que se requiere corriente para que brille la luz. La batería tiene corriente eléctrica almacenada que permite que la luz brille hasta que la corriente eléctrica esté presente en la batería. Esta batería se puede cargar durante un período de tiempo cuando está llena, se puede recuperar durante muchas horas.

Answer:

a) A = 3 cm,  b)  T = 0.4 s,   f = 2.5 Hz,

2) A standing wave the displacement of the wave is canceled and only one oscillation remains

Explanation:

a) in an oscillatory movement the amplitude is the highest value of the signal in this case

          A = 3 cm

b) the period of oscillation is the time it takes for the wave to repeat itself in this case

          T = 0.4 s

the period is the inverse of the frequency

         f = 1 /T

         f = 1 /, 0.4

         f = 2.5 Hz

2) a traveling wave is a wave for which as time increases the displacement increases, in the case of a transverse wave the oscillation is perpendicular to the displacement and in the case of a longitudinal wave the oscillation is in the same direction of the displacement.

A standing wave occurs when a traveling wave bounces off some object and there are two waves, one that travels in one direction and the other that travels in the opposite direction. In this case, the displacement of the wave is canceled and only one oscillation remains.

Answer:

The force is 35808 N.

Explanation:

Power, P = 600 hp = 600 x 746 W = 447600 W

speed, v = 45 km/h = 12.5 m/s

Let the force is F.

P = F v

447600 = F x 12.5

F = 35808 N

Answer:

V = 50 volts

Explanation:

Given that,

Resistance, R = 10 ohms

Current, I = 5 A

We need to find the potential difference across the circuit. We know that,

V = IR

Put all the values,

V = 5 × 10

V = 50 volts

Hence, the potential difference is equal to 50 volts.

Answer:

Density=Mass/Volume

            =20/4 kg/m^3

            =5 kg/m^3

Answer:

0.3 cm

Explanation:

[tex] \frac{2.4}{8} \\ = 0.3[/tex]

Answer:

Answer:It is due to tyndall effect that the path of a light ray passing through a suspension is visible. The particles in suspension scatter the beam of light making it clearly visible.

Answer:

it is due to the Tyndall effect so the path of light is visible in a suspension but in solution the path of light is not visible

Answer:

Stimulants are a group of drugs that result in increased activity in the body. Sometimes referred to as “uppers,” these drugs are frequently abused due to their performance-enhancing and euphoric effects. Generally, those who abuse stimulants experience heightened energy levels and enhanced focus.

Stimulants speed up mental and physical processes, which can produce desirable effects in the short-term by increasing levels of dopamine in the brain. While users may feel great due to the short-term effects of stimulants, long-term abuse of these drugs can have significant consequences, which is why it is so important for those who abuse the drugs to get help as quickly as possible.

Explanation:

hope this works

Answer:

his displacement is 772.85 ft

Explanation:

Given;

initial velocity of his jump, u = 2 ft/s

final velocity of his jump, v = - 223 ft/s

time of motion, t = 7 seconds

acceleration due to gravity, g = 32.17 ft/s²

Let downward motion = positive direction

Let his displacement after 7s = Δh

Apply the following kinematic equation to determine his displacement.

[tex]v^2 = u^2 + 2g\Delta h\\\\(-223)^2 = (2)^2 + (2\times 32.17)\Delta h\\\\49,729 = 4 + 64.34\Delta h\\\\-64.34 \Delta h = 4 - 49,729\\\\-64.34 \Delta h = -49,725\\\\\Delta h = \frac{49,725}{64.34} \\\\\Delta h = 772.85 \ ft[/tex]

Therefore, his displacement is 772.85 ft

Answer:

El peso de la persona B es la mitad del peso de la persona A.

Explanation:

El peso de la persona B puede calcularse con la siguiente ecuación:

[tex] P_{B} = m_{B}g [/tex]   (1)

En donde:

[tex]m_{B}[/tex]: es la masa de la persona B

g: es la gravedad

Dado que la persona B tiene la mitad de la masa de la persona A, tenemos:

[tex] m_{B} = \frac{m_{A}}{2} [/tex]  (2)

En donde:

[tex]m_{A}[/tex]: es la masa de la persona A

Al introducir la ecuación (2) en (1) nos queda:

[tex] P_{B} = \frac{m_{A}}{2}g [/tex]   (3)

Sabemos que el peso de la persona A está dado por:

[tex] P_{A} = m_{A}g [/tex]   (4)

Entonces, al introducir la ecuación (4) en (3) tenemos:

[tex] P_{B} = \frac{P_{A}}{2} [/tex]

Por lo tanto, el peso de la persona B es la mitad del peso de la persona A.

Espero que te sea de utilidad!

Answer:

Por que existe una relación entre dichas magnitudes.

Para un pendulo perfecto de largo L, sabemos que el periodo esta definido como:

[tex]T = 2*\pi *\sqrt{L/g}[/tex]

Donde:

pi = 3.14

L = largo del péndulo

g = aceleración gravitatoria = 9.8 m/s^2

Entonces podemos ver que el periodo es proporcional a la raíz cuadrada de la longitud del péndulo.

Esto es por lo que al analizar datos de un experimento se debería observar una relación entre esas magnitudes.

Answer:

When a current-carrying coil is suspended in a uniform magnetic field it is acted upon by a torque. Under the action of this torque, the coil rotates and the deflection in the coil in a moving coil galvanometer is directly proportional to the current flowing through the coil.