Answer:
8.4 N/m
Explanation:
m = Mass of block = 4.63 gm
g = Acceleration due to gravity = 
x = Displacement of spring = 0.45 cm
a = Acceleration of subject = 0.832g
k = Spring constant
Force is given by
From Hooke's law
So
The force constant of the spring is 8.4 N/m.
Answer:
Force from the support closest = 79.8 N
Force from the support furthest = 61.9 N
Explanation:
Let's say the length of the beam is L. Let's say A is the near end of the beam and B is the far end of the beam.
Draw a free body diagram. There are four forces on the beam:
Reaction force Ra at the near end (0),
Reaction force Rb at the far end (L),
Weight force of the beam Mg at the center (L/2),
Weight force of the book mg at L/4 from A.
Sum of torques at A:
∑τ = Iα
Rb (L) − Mg (L/2) − mg (L/4) = 0
Rb (L) = Mg (L/2) + mg (L/4)
Rb = ½ Mg + ¼ mg
Rb = (½ M + ¼ m) g
Rb = (½ (10.8 kg) + ¼ (3.66 kg)) (9.8 m/s²)
Rb = 61.9 N
Sum of forces in the y direction:
∑F = ma
Ra + Rb − Mg − mg = 0
Ra = Mg + mg − Rb
Ra = (M + m) g − Rb
Ra = (10.8 kg + 3.66 kg) (9.8 m/s²) − 61.9 N
Ra = 79.8 N
Place mixtuer into water. stir mixture into water, wait untill salt is disoldved. Strain out grain. boil water intill only salt is left. Hope this helps! ^-^
<span>Depth = 5.0 Ă— 10^2 m
Density of sea water = 1.025 x 10^3
Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa
Since the normal pressure is retained in the hull, no need to bother about Po
Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
So it is 49.56 times larger.</span>
Acceleration no longer exist as the car stops.
