Answer: 117.60N
Explanation:
Weight is a force. Therefore, we can use the force formula to find weight.

W = weight
m = mass
g = acceleration due to gravity (
)

Answer:
heterogeneous mixture has components that are not evenly distributed. This means that you can easily distinguish between the different components.
The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × 
friction factor = 0. 52 × 
friction factor = 0. 52 × 0. 55
friction factor 
b. When V = 3mls
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 0. 185
Friction factor 
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 ×
×
× 
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss =
×
×
× 
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
Learn more about friction here:
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Answer:
Explanation:
initial angular velocity, ωo = 0 rad/s
angular acceleration, α = 30.5 rad/s²
time, t = 9 s
radius, r = 0.120 m
let the velocity is v after time 9 s.
Use first equation of motion for rotational motion
ω = ωo + αt
ω = 0 + 30.5 x 9
ω = 274.5 rad/s
v = rω
v = 0.120 x 274.5
v = 32.94 m/s
When copper carbonate is heated, it decomposes to form a black residue. This black residue is actually copper(II) oxide. Along with this black residue, carbon-dioxide is released. The complete reaction is the following:

The law of conservation of mass holds true. 123.6 g of copper carbonate decomposes to 79.6 g of copper oxide and (123.6 g-79.6 g=44 g) 44 g of carbon-dioxide.