Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
1. 1 s = 1 x 10⁶ μs
2. 1 g = 0.001 kg
3. 1 km = 1000 m
4. 1 mm = 1 x 10⁻³ m
5. 1 mL = 1 x 10⁻³ L
6. 1 g = 100 dg
7. 1 cm = 1 x 10⁻² m
8. 1 ms = 1 x 10⁻³ s
Explanation:
1.
1 x 10⁻⁶ s = 1 μs
(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs
<u>1 s = 1 x 10⁶ μs</u>
2.
1000 g = 1 kg
1 g = 1/1000 kg
<u>1 g = 0.001 kg</u>
3.
<u>1 km = 1000 m</u>
<u></u>
4.
<u>1 mm = 1 x 10⁻³ m</u>
<u></u>
5.
<u>1 mL = 1 x 10⁻³ L</u>
<u></u>
6.
1 x 10⁻² g = 1 dg
(1 x 10⁻² x 10²) g = 1 x 10² dg
<u>1 g = 100 dg</u>
<u></u>
7.
<u>1 cm = 1 x 10⁻² m</u>
<u></u>
8.
<u>1 ms = 1 x 10⁻³ s</u>
Answer:
<em>The centripetal acceleration would increase by a factor of 4</em>
<em>Correct choice: B.</em>
Explanation:
<u>Circular Motion</u>
The circular motion is described when an object rotates about a fixed point called center. The distance from the object to the center is the radius. There are other magnitudes in the circular motion like the angular speed, tangent speed, and centripetal acceleration. The formulas are:
If the speed is doubled and the radius is the same, then
The centripetal acceleration would increase by a factor of 4
Correct choice: B.
Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2
Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by
where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is
Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,
Therefore, the additional work needed is
Learn more about work:
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