Answer:
40 metros en 10 segundos.
Explicación:
Una grúa está levantando una caja de 1000 kg atada a una cadena y la caja, que inicialmente está en reposo, aumenta su velocidad en 4 m / s por segundo, por lo que si la región de colocación está a 40 metros de distancia, la grúa tarda 10 segundos en completar el proceso. tarea de movimiento y colocación de la caja. La grúa se utiliza para levantar y mover cargas pesadas, máquinas, materiales y mercancías para diferentes propósitos. Entonces el trabajo realizado por la grúa depende de la velocidad y la distancia de colocación.
The mass of the man would remain the same…
His weight would change. Assuming the man’s weight on earth is 60N.
Since weight is a force, according to Newton’s second law, F = ma ( m = mass, a = acceleration due to gravity) First lets find the mass of the man, as it is required to find his weight on the moon.
F=ma[taking a of earth as10m/s
2
]
60=m.10[divide10onbothsides]
m=
10
60
= 6 Kg
Acceleration due to gravity on the moon is83%less than the acceleration due to gravity on earth(1.622m/s
2
).
F=ma
F=6.1.622=9.732 N
So a person weighting 60N on earth would approximately weight around 10Non the moon.
Answer: 2
Explanation:
Each shell can contain only a fixed number of electrons: The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. The general formula is that the nth shell can in principle hold up to 2(n2) electrons.
<span>10.7 hours
Since we've been told that we have a speed of 1 km/h and a width of 0.5 meters that means that after 1 hour, we can mow a path 0.5 m wide and 1000 m long for a total area of 1000 m * 0.5 m = 500 m^2/hr. Now let's calculate the area of that football field.
360 ft / 3.281 ft/m * 160 ft / 3.281 ft/m = 5350.692864 m^2
Now let's divide that area by the 500 m^2 calculated earlier to see how many hours it will take.
5350.692864 m^2 / 500 m^2/hr = 10.70138573 hr
So it would take about 10.7 hours to mow the football field.</span>
