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Anvisha [2.4K]
2 years ago
7

4. Un móvil viaja en línea recta con una velocidad media de 1200 cm/s durante 9 s, y luego con velocidad media de 480 cm/s duran

te 7 s, siendo ambas velocidades del mismo sentido: a) ¿cuál es el desplazamiento total en el viaje de 16 s? B) ¿cuál es la velocidad media del viaje completo?
Physics
1 answer:
katrin [286]2 years ago
7 0

Answer:

A) El desplazamiento total del viaje es 14160 centímetros.

B) La velocidad media del viaje completo es 885 centímetros por segundo.

Explanation:

A) El desplazamiento es el producto de la velocidad media por el tiempo. El desplazamiento total es la suma de desplazamientos asociados a cada velocidad media, entonces:

x = \left(1200\,\frac{cm}{s}\right)\cdot (9\,s)+ \left(480\,\frac{cm}{s} \right)\cdot (7\,s)

x = 14160\,cm

El desplazamiento total del viaje es 14160 centímetros.

B) La velocidad media del viaje es el desplazamiento total dividido por el tiempo total, es decir:

\bar v = \frac{14160\,cm }{16\,s}

\bar v = 885\,\frac{cm}{s}

La velocidad media del viaje completo es 885 centímetros por segundo.

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Can an object be accelerated while traveling at constant velocity? Why or why not?
AysviL [449]

Answer:

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.

3 0
3 years ago
PE=?, m=.6kg, g=10m/s2, h=35m<br><br> PLS HELP I NEED THIS DONE
Karo-lina-s [1.5K]
210J

PE is mgh in this context.
7 0
3 years ago
Two objects attract each other gravitationally. If the distance between their centers decreases by a factor of 2, how does the g
kramer

Answer:

The gravitational force between them increases by a factor of 4

Explanation:

Gravitational force is a force of attraction between two objects with masses M and m which are separated by a distance R. It is given mathematically as:

Fg = GMm/R²

Where G = Gravitational constant.

If the distance between their centers, R, decreases by a factor of 2, then it means the new distance between their centers is:

r = R/2

Hence,the gravitational force becomes:

Fg = GMm/r²

Fg = GMm/(R/2)²

Fg = GMm/(R²/4)

Fg = 4GMm/R²

Hence,the gravitational force increases by a factor of 4.

6 0
3 years ago
A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.
Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0
3 years ago
What instrument is used to expand burr holes?
k0ka [10]
Craniotomes are used
4 0
3 years ago
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