We have: K.E. = mv² / 2
Here, m = 1 Kg
v = 4 m/s
Substitute their values in the formula,
K.E. = 1×4² / 2
K.E. = 16/ 2
K.E. = 8 J
Finally, answer of your question would be 8 Joules.
Hope this helps!
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate doesn't move (static), acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = Fm - Ff = 0.
Fm is the applied force
Ff is the frictional force
Since Fm - Ff = 0
Fm = Ff
This means that the applied force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
R = 31.2 × 9.8
R = 305.76N
From the formula
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
Answer: 0.24g/ml
Explanation:
Given that:
Volume of water displaced = 23.5 ml
Mass of cork = 5.7 g
Density of the cork = ?
Recall that density is obtained by dividing the mass of a substance by the volume of water displaced.
i.e Density = Mass/volume
Density = 5.7g /23.5ml
Density = 0.24g/ml
Thus, the density of the piece of cork is 0.24g/ml
Answer:
a) m = 59.63 [kg]
b) Wm = 95.41 [N]
Explanation:
El peso de un cuerpo se define como el producto de la masa por la aceleración gravitacional. DE esta manera tenemos:
W = m*g
Donde:
m = masa [kg]
g = gravedad = 9.81 [m/s^2]
m = W / g
m = 585 / 9.81
m = 59.63 [kg]
Es importante aclarar que la masa se conserva independientemente de la ubicación del cuerpo en el espacio.
Por ende su masa sera la misma en la luna.
El peso en la luna se calcula como Wm y es igual a:
Wm = 59.63 * 1.6 = 95.41 [N]
