Answer:
1,85 m / s²
Explanation:
De la pregunta anterior, se obtuvieron los siguientes datos:
Velocidad inicial (u) = 40 km / h
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:
1 km / h = 0,2778 m / s
Por lo tanto,
40 km / h = 40 km / h × 0,2778 m / s / 1 km / h
40 km / h = 11,11 m / s
Por tanto, 40 km / h equivalen a 11,11 m / s.
Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:
Velocidad inicial (u) = 11,11 m / s
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
a = (v - u) / (t₂ - t₁)
a = (0 - 11,11) / (6 - 0)
a = - 11,11 / 6
a = –1,85 m / s²
Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²
Here it is an application of Newton's III law
as we know by Newton's III law that every action has equal and opposite reaction
So here as we know that two boys jumps off the boat with different forces
from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude
So boat must have an opposite force on front end with magnitude 45 N
Now similar way we can say
from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude
So boat must have an opposite force on front end with magnitude 60 N
So here net force due to both jump on the boat is given by



so boat will have net force F = 15 N in forward direction due to both jumps
-- <u><em>Current is measured in amps.</em></u> (You can use any symbol you want to represent current, but the most common one is " I ", not "Δ".)
-- <u><em>The relationship between current, voltage, and resistance is mathematically defined by Ohm's Law. </em></u>
-- <u><em>Current is the flow of electrons through a circuit.</em></u>
-- (Ohm's Law is NOT mathematically represented by the equation V=I/R.) <u><em>It should be V = I · R</em></u> .
(When solving for Resistance in a circuit and both voltage and current are known values, the equation I =V*R is not true, and not the way to solve it.) <u><em>If the resistance is what you're looking for, then the equation to use is </em></u><u><em>R = V / I</em></u><u><em> . </em></u>
<em>-- </em><u><em>If the voltage in a circuit is increased, the current will also increase.</em></u>
The intensity of the light has no connection with the photoelectric effect.
That's what was so baffling about it before the particle nature of light
was suspected ... a match with a blue flame might stimulate the
photoelectric effect, but a high-power red searchlight couldn't do it.
Current = (voltage) / (resistance)
= (35 volts) / (1,400 ohms)
= 25 milliamperes