A. The magnitude of the spring force (in N) acting upon the object is 15.9 N
B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²
C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).
<h3>A. How to determine the force </h3>
- Extension (e) = 0.150 m
- Spring constant (K) = 106 N/m
- Force (F) = ?
F = Ke
F = 106 × 0.15
F = 15.9 N
<h3>B. How to determine the acceleration</h3>
- Mass (m) = 0.52 Kg
- Force (F) = 15. 9 N
- Acceleration (a) =?
F = ma
Divide both sides by m
a = F / m
a = 15.9 / 0.52
a = 30.58 m/s²
<h3>C. How to determine the direction of the acceleration vector</h3>
Considering the diagram, we can see that the spring was pulled away from the equilibrium point.
Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.
Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.
Learn more about spring constant:
brainly.com/question/9199238
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Answer:
ΔP = 20000 N s
Explanation:
To solve this problem we use the relation between momentum and moment
I = Δp
let's calculate the momentum
I = ∫F dt
if we use the average force
I = F t
I = 10000 2
I = 20000 N s
therefore with the first equation
ΔP = I = 20000 N s
F = ma
Rearrange this so acceleration is the subject:
A = f/m
Now input your values into this equation for the first question :
A = f/m
A = 15/45
A = 0.33333333 m/s
Then using this same equation change the force of 15 to 25 :
A = f/m
A = 25/45
A = 0.55555556 m/s
For the last question keep the force of 15N but change the mass to 70kg into the same equation :
A = f/m
A = 15/70
A = 0.21 m/s (rounded)
Answer:
A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)
Explanation:
<u>To rank the angular speed (ω) of the objects, we need first calculate its value for every object:</u>
A bowling ball of radius 12.3cm rotating at 8.21 radians per second:
ω = 8.21 rad/s
A tire of radius 0.321m rotating at 75.8 rpm:
A 6.84cm diameter top spinning at 375 degrees per second:
A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:
A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:
<u>Now, the rank of the angular speed of the objects, from highest to lowest is: </u>
A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)
I hope it helps you!
