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3 years ago

The average speed of a runner in a 400 meter race was 8.0 meters per second how long did it take the runner to complete the race

2 answers:

8 0

If it is eight meters per second, and the total distance he has to travel is four hundred meters to complete the race, you divide the total length needed to travel by the average speed per second to see the amount of seconds required to complete the race at that speed.

8 meters per second: Speed of runner
400 meters: Total distance

400/8=50

50 seconds to complete the race at the eight of 8m/s.

erik [133]3 years ago

7 0

Answer:

If it is eight meters per second, and the total distance he has to travel is four hundred meters to complete the race, you divide the total length needed to travel by the average speed per second to see the amount of seconds required to complete the race at that speed.8 meters per second: Speed of runner400 meters: Total distance400/8=50.50 seconds to complete the race at the eight of 8m/s.

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A ball is thrown vertically upwards. It returns 6s later. Calculate : (1) the greatest height reached by the ball, and (2) the i

dsp73

Answer:

greatest displacement = 44.1m

initial velocity= 29.4m/s

Explanation:

Greatest displacement

s=1/2at^2

= (9.8/2 ×9)m

= 44.1m

initial velocity

s=ut-1/2at^2

44.1= 3u -(1/2×9.8×9)

44.1=3u-44.1

3u=88.2

u=29.4m/s

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4 years ago

Un cuerpo se lanza verticalmente hacia arriba con una velocidad de 13 m/s. ¿Cuánto tiempo tarda en alcanzar la altura máxima? a)

Elena L [17]

Answer:

D. 1.33 segundos.

Explanation:

El cuerpo es experimenta un movimiento en caída libre al modificarse su velocidad por efecto de la gravitación terrestre. Este cuerpo alcanza instantáneamente el reposo cuando se encuentra a su altura máxima, el tiempo puede obtenerse sabiendo la aceleración y las velocidades incial y final a partir de la siguiente ecuación cinemática:

$v = v_{o}+g\cdot t$

Donde:

$v$ - Velocidad final del cuerpo, medida en metros por segundo.

$v_{o}$ - Velocidad inicial del cuerpo, medida en metros por segundo.

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

$t$ - Tiempo, medido en segundos.

Ahora se despeja el tiempo:

$t = \frac{v-v_{o}}{g}$

Si $v_{o} = 13\,\frac{m}{s}$ , $v=0\,\frac{m}{s}$ y $g = -9.807\,\frac{m}{s^{2}}$ , entonces:

$t = \frac{0\,\frac{m}{s}-13\,\frac{m}{s}}{-9.807\,\frac{m}{s^{2}} }$

$t = 1.326\,s$

Por ende, la respuesta correcta es D.

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3 years ago

O que é fisiológicos na areia medica

Makovka662 [10]

??????????????????????????

7 0

2 years ago

Ben Rushin is waiting at a stop light. Turns green, ben accelerated from rest at a rate of 6.00 m/s squared for a time of 4.10 s

Lera25 [3.4K]

D= vt +.5at^2
since he started at rest, v (initial velocity) is 0
so d=.5at^2
d = .5 (6m/s^2) (4.1s)^2
then put that into a calculator.

4 0

3 years ago

The three forces shown act on a particle. what is the direction of the resultant of these three forces?

melisa1 [442]

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
$F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N$
in the positive direction.

y-axis) The resultant on the y-axis is
$F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N$
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
$\tan \alpha = \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7$
from which we find
$\alpha=35^{\circ}$

7 0

3 years ago