Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
Density is defined as (mass) per unit (volume). So in order to calculate
the density of a glob of some substance, you pretty much have to measure
its mass and its volume.
Answer:
Explanation:
Work
Other units Foot-pound, Erg
In SI base units 1 kg⋅m2⋅s−2
Derivations from other quantities W = F ⋅ s W = τ θ
Dimension M L2 T−2
Idk if this is what u are looking for but i hope this help.:)
We first determine the vertex by using the formula,<span>-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got
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v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948
then
let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is
23/15.948=1.44218 sec
substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329
determine vertex by using a graphing calculatort=2.1594s h=22.85m
using the time value of the vertex, determine horizontal distance travelled
34.438m away from cannon
Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
