Answer:
El coeficiente de fricción entre el bloque y el piso es 0,64.
Explanation:
Podemos hallar el coeficiente de fricción usando la definición de trabajo:
En donde:
Kf es la energía cinética final =0
Ki es la energía cinética inicial
m: es la masa = 2 kg
v: es la velocidad = 5 m/s
Además el trabajo es el producto de la fuerza de rozamiento (Fr) por el desplazamiento (d), por lo que tenemos:
Por lo tanto, el coeficiente de fricción entre el bloque y el piso es 0,64.
Espero que te sea de utilidad!
Answer:
(a) Total north component = (26.89 + 25) = 51.89 metres
(b) Total west component = (-34.83 + 52) = 17.17 metres.
Explanation:
Do northwest components first.
North component = (sin 33.9 x 44) =26.89 metres.
West component = (cos 33.9 x 44) = -34.83metres.
Total west component = (-34.83 + 52) = 17.17 metres.
Total north component = (26.89 + 25) = 51.89 metres
Answer:
Velocity of truck will be 20.287 m /sec
Explanation:
We have given mass of the truck m = 4000 kg
Radius of the turn r = 70 m
Coefficient of friction
Centripetal force is given
And frictional force is equal to
For body to be move these two forces must be equal
So
Answer:
the potential energy of this body is 245 J.
Explanation:
Given;
mass of the body, m = 250 g = 0.25 kg
height from which the body was dropped, h = 100 m
acceleration due to gravity, g = 9.8 m/s²
The potential energy of this body is calculated as;
P.E = mgh
substitute the given values and solve for the potential energy of this body;
P.E = 0.25 x 9.8 x 100
P.E = 245 J.
Therefore, the potential energy of this body is 245 J.
<span>KE = 1/2 * m * v^2 = 1/2 * 30 * 20^2 = 1/2 * 30 * 400 = 6000 J</span>