Answer:
a) N = 9 Mg
, b)N_w = μ 9M
, c)
Explanation:
a) For this part we write the equations of trslacinal equilibrium
Axis y
N - Mg - 8M g = 0
N = 9 Mg
N = 9 11 9.8
N = 970.2 N
b) the force on the horizontal axis (x) som
fr -N_w = 0
fr = N_w
friction force is
fr = μ N
N_w = μ 9M
g
fr = 0.59 970.2
fr = N_w = 572,418 N
c) For this part we must use rotational equilibrium.
Στ = 0
We set a frame of reference at the bottom of the ladder and assume that the counterclockwise acceleration is positive
the weight of it is at its midpoint (L / 2)
- W L /2 cos 54 - 8M d_max cos 54+ NW L sin 54 = 0
8M d_max cos 54 = - W L / 2 cos 54 + NW L sin 54
d_max = L (-Mg 1/2 cos 54 + NW sin 54) / (8M cos 54)
d_max = L (-g / 16 + μ 9Mg / 8M tan 54)
d_max = L ( 9/8 μ g tan 54- g/16)
It's C. increasing the masses of the objects and decreasing the distance between the objects
Answer:
CaCl2 is a reactant
Explanation:
Calcium carbonate (CaCO3) is a reagent, it is found on the left side, what is found on the right side are the products.
Answer:
the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Explanation:
a) Kinetic energy of block = potential energy in spring
½ mv² = ½ kx²
Here m stands for combined mass (block + bullet),
which is just 1 kg. Spring constant k is unknown, but you can find it from given data:
k = 0.75 N / 0.25 cm
= 3 N/cm, or 300 N/m.
From the energy equation above, solve for v,
v = v √(k/m)
= 0.15 √(300/1)
= 2.598 m/s.
b) Momentum before impact = momentum after impact.
Since m = 1 kg,
v = 2.598 m/s,
p = 2.598 kg m/s.
This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,
u = 2.598 kg m/s / 8 × 10⁻³ kg
= 324.76 m/s.
Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
The amplitude of the wave is the 'full height of the wave.' Amplitude is measured in m (meters) and is measured over the change of a single period.
