Answer:
The new speed of the ball is 176.43 m/s
Explanation:
Given;
mass of the ball, m = 7 kg
initial speed of the ball, u = 5 m/s
applied force, F = 300 N
time of force action on the ball, t = 4 s
Apply Newton's second law of motion;

where;
v is new speed of the ball

Therefore, the new speed of the ball is 176.43 m/s
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hope this helps, Terrakmiller80! :-)</span>
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.
If the ration supplementary angle is 11:7,find the measure of the larger angle larger angle?