Answer:m1v1 + m2v2 = (m1f + m2f)vf. 3000kg(10.0m/s) + (15000kg)(0.0m/s) = (18000kg)(vf).
Explanation:
Answer:
C. At the bottom of the circle.
Explanation:
Lets take
Radius of the circle = r
Mass = m
Tension = T
Angular speed = ω
The radial acceleration towards = a
a= ω² r
Weight due to gravity = mg
<h3>At the bottom condition</h3>
T - m g = m a
T = m ω² r + m g
<h3>At the top condition</h3>
T + m g = m a
T= m ω² r -m g
From above equation we can say that tension is grater when ball at bottom of the vertical circle.
Therefore the answer is C.
C. At the bottom of the circle.
Answer:5.7m/s
Explanation:
Mass=1kg
Initial velocity=u=8m/s
height=h=1.6m
Final velocity =v
Acceleration due to gravity=g=9.8m/s^2
v^2=u^2-2xgxh
v^2=8^2-2x9.8x1.6
v^2=8x8-2x9.8x1.6
v^2=64-31.36
v^2=32.64
Take the square root of both sides
√(v^2)=√(32.64)
v=5.7
Speed at the height of 1.6m is 5.7m/s
Do not approach within 100-yards, and slow to minimum speed within 500-yards of any U.S. naval vessel.
Answer:
The maximum speed is 21.39 m/s.
Explanation:
Given;
radius of the flat curve, r₁ = 150 m
maximum speed,
= 32.5 m/s
The maximum acceleration on the unbanked curve is calculated as;

the radius of the second flat curve, r₂ = 65.0 m
the maximum speed this unbanked curve should be rated is calculated as;

Therefore, the maximum speed is 21.39 m/s.