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rosijanka [135]
3 years ago
7

Which is bigger, a kilometer or a mile?

Physics
2 answers:
Marta_Voda [28]3 years ago
7 0
Mile is much more bigger than kilometer.
I am Lyosha [343]3 years ago
7 0
1 mile = approx 1.609 kilometer
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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2 . The acc
Eddi Din [679]

Answer:

The maximum height reached by the rocket is 1.94 × 10³ m.

Explanation:

The height of the rocket can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²    (when the rocket is accelerated upward).

y = y0 +  v0 · t + 1/2 · g · t² (after the rocket runs out of fuel).

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to engines of the rocket.

g = acceleration due to gravity.

In the same way, the velocity of the rocket can be calculated as follows:

v = v0 + a · t  (when the rocket has fuel)

v = v0 + g · t   (when the rocket runs out of fuel)

Where "v" is the velocity at time "t"

First, let´s find the height reached until the rocket runs out of fuel.

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · t + 1/2 · 34.3 m/s² · (5.00 s)²

y = 429 m

And now, let´s find the velocity reached in that time of upward acceleration:

v = v0 + a · t

v = 0 m/s + 34.3 m/s² · 5.00 s

v = 172 m/s

When the rocket runs out of fuel, it is accelerated downward due to gravity. But, since the rocket has initially an upward velocity (172 m/s), it will not fall immediately and will continue to go up until the velocity becomes 0. In that instant, the rocket is at its maximum height and thereafter it will start to fall with negative velocity.

Then, using the equation for velocity, we can calculate the time it takes the rocket to reach its maximum height:

v = v0 + g · t

0 = 172 m/s - 9.80 m/s² · t

-172 m/s / -9.80 m/s² = t

t = 17.6 s

With this time, we can now calcualte the maximum height. Notice that the initial velocity and height are the ones reached during the upward acceleration phase:

y = y0 +  v0 · t + 1/2 · g · t²

ymax = 429 m + 172 m/s · 17.6 s - 1/2 · 9.80 m/s² · (17.6 s)²

ymax = 1.94 × 10³ m

4 0
4 years ago
Erosion piles up in low places to form:
beks73 [17]
Soil and sand
Erosion is the movement of sand or sand particles caused by the fires of the wind
7 0
3 years ago
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An object is moving at 4.66 m/s when it accelerates at 5.66 m/s2 for a period of 2.35 s. The distance it moves in this time is _
valentinak56 [21]

Answer:

S = 26.58 meters

Explanation:

Given the following data;

Initial velocity = 4.66 m/s

Acceleration = 5.66 m/s²

Time = 2.35 seconds

To find the distance travelled by the object, we would use the second equation of motion;

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

S = 4.66*2.35 + ½*5.66*2.35²

S = 10.951 + (2.83 * 5.5225)

S = 10.951 + 15.629

S = 26.58 meters

6 0
3 years ago
Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente h
shutvik [7]

Answer:

a) La piedra es lanzada desde una altura de 0,785 metros.

b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

Explanation:

a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:

y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2} (1)

Donde:

y - Altura final, medida en metros.

y_{o} - Altura inicial, medida en metros.

v_{o,y} - Componente vertical de la velocidad inicial, medida en metros por segundo.

t - Tiempo, medido en segundos.

g - Aceleración gravitacional, medida en metros por segundo cuadrado.

Si sabemos que y = 0\,m, v_{o,y} = 0\,\frac{m}{s}, t = 0,4\,s y g = -9,807\,\frac{m}{s^{2}}, entonces la altura inicial de la piedra es:

y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}

y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}

y_{o} = 0,785\,m

La piedra es lanzada desde una altura de 0,785 metros.

b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:

v_{o,x} = \frac{x-x_{o}}{t} (2)

Donde:

x_{o}, x - Posiciones horizontales iniciales y finales, medidas en metros.

t - Tiempo, medido en segundos.

Si tenemos que x_{o} = 0\,m, x = 2,5\,m y t = 0,4\,s, entonces el componente horizontal de la velocidad inicial es:

v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}

v_{o,x} = 6,25\,\frac{m}{s}

La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

4 0
3 years ago
Hydrogen (H-1), deuterium (H-2), and tritium (H-3) are the three isotopes of hydrogen. What are the values of a, b, and c in the
asambeis [7]

I believe it is a=1 b=2 c=3

plato students

6 0
3 years ago
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