The power developed is 500 W ( to the nearest Watt)
Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.


Therefore,

Substitute the given values of force , displacement and time


Thus the Power can be rounded off to the nearest value of 500 W
Answer:
i) 24.5 m/s
ii) 30,656 m
iii) 89,344 m
Explanation:
Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer
i) Los parámetros dados son;
Altura inicial, s = 120 m
El tiempo en caída libre = 2.5 s
De la ecuación de caída libre, tenemos;
v = u + gt
Dónde:
u = Velocidad inicial = 0 m / s
g = Aceleración debida a la gravedad = 9.81 m / s²
t = Tiempo de caída libre = 2.5 s
Por lo tanto;
v = 0 + 9.8 × 2.5 = 24.5 m / s
ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación
s = u · t + 1/2 · g · t²
= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m
iii) La altura restante = 120 - 30.656 = 89.344 m.
Explanation:
Power of electric kettle, P = 1 kW
Voltage, V = 220 V
(a) Electric power is given by the formula as follows :

R is resistance

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.
Energy supplied is given by :

P is power, 

Answer:
it depends on a person's own weight
Explanation:
It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².
The second equation of kinematics gives the relationship between the height reached and time taken by it.
Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.
We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

t is time taken by the ball to hit the ground
is initial speed of the ball
So, the correct option is (A).