Answer:
Team A wins the frame with three points. The opposing team gets zero points for the frame.
Explanation:
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
<h3>Answer</h3>
6.6 N pointing to the right
<h3>Explanation</h3>
Given that,
two forces acting of magnitude 3.6N
angle between them = 48°
To find,
the third force that will cause the object to be in equilibrium
<h3>1)</h3>
Find the vertical and horizontal components of the two forces
vertical force1 = sin(24)(3.6)
vertical force2= -sin(24)(3.6)
<em>(negative sign since it is acting on opposite direction)</em>
vertical force3 = sin(24)(3.6) - sin(24)(3.6)
= 0
<h3>2)</h3>
horizontal force1 = cos(24)(3.6)
horizontal force2= cos(24)(3.6)
horizontal force3 = cos(24)(3.6) + cos(24)(3.6)
= 2(cos(24)(3.6))
= 6.5775 N
≈ 6.6 N
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