Answer:
tan is 15 for that triangle
Answer:
3141N or 3.1 ×10³N to 2 significant figures. The can experiences this inward force on its outer surface.
Explanation:
The atmospheric pressure acts on the outer surface of the can. In order to calculate this inward force we need to know the total surface area of the can available to the air outside the can. Since the can is a cylinder with a total surface area given by 2πrh + 2πr² =
A = 2πr(r + h)
Where h = height of the can = 12cm
r = radius of the can = 6.5cm/2 = 3.25cm
r = diameter /2
A = 2π×3.25 ×(3.25 + 12) = 311.4cm² = 311.4 ×10-⁴ = 0.031m²
Atmospheric pressure, P = 101325Pa = 101325 N/m²
F = P × A
F = 101325 ×0.031.
F = 3141N. Or 3.1 ×10³ N.
Answer:
0.219N
Explanation:
Given data
mass= 7.3kg
acceleration= 0.03m/s^2
We know that
F=ma
substitute
F= 7.3*0.03
F= 0.219N
Hence the applied force is 0.219N
I think an object such as an iron pipe could work
