Answer:
1.65
Explanation:
The equation of the forces along the horizontal direction is:
(1)
where
F = 65 N is the force applied with the push
is the frictional force
m = 4 kg is the mass
is the acceleration
The force of friction can be written as 
(2), where
is the coefficient of kinetic friction
R is the normal force exerted by the floor
The equation of forces along the vertical direction is
(3)
since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find
And solving for ,
Answer:
714.285s
Explanation:
use relative velocity
8-4.5 = 3.5m/s
x = 2500m
2500/3.5 = 714.285s = 700s (with sig figs)
W=20 e(-kt)
A. Rearranging gives k= -(ln(w/20)/t
Substituting w= 10 and solving gives k=0.014
B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g
C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours.
D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h
E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW
Answer:
The distance covered by the sprinter, s = 40 m
Explanation:
Given data,
The initial velocity of the sprinter, u = 0 m/s
The acceleration of the sprinter, a = 5 m/s²
The time period of acceleration of the sprinter, t = 4 s
Using the II equations of motion
s = ut + ½ at²
= 0 + ½ (5) (4)²
= 40 m
Hence, the distance covered by the sprinter, s = 40 m
