The net force on the block parallel to the incline is
∑ F = -mg sin(θ) = ma
where m is the mass of the block, g = 9.8 m/s² is the acceleration due to gravity, θ is the angle the incline makes with the horizontal, and a is the acceleration of the block. Solving for a gives
a = -g sin(θ)
so the block would slide down the incline with acceleration
a = - (9.8 m/s²) sin(30°) = -4.9 m/s²
Answer:
a) Weight of the rock out of the water = 16.37 N
b) Buoyancy force = 4.61 N
c) Mass of the water displaced = 0.47 kg
d) Weight of rock under water = 11.76 N
Explanation:
a) Mass of the rock out of the water = Volume x Density
Volume = 470 cm³
Density = 3.55 g/cm³
Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg
Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N
b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.
Volume = 470 cm³
Density of liquid = 1 g/cm³

c) Mass of the water displaced = Volume of body x Density of liquid
Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg
d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force
Weight of rock under water = 16.37 - 4.61 =11.76 N
Answer:
0.43 s
Explanation:
We have the following parameters:
Initial velocity, u = 7.4 m/s
Acceleration of gravity, g = 9.8 
Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m
Time, t = ?
Using the equation of motion
, we have


Using the quadratic formula
where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have
s
Answer. dry friction- <span> resists relative lateral motion of two solid surfaces in contact.</span>