Asked by he
What is the acceleration of the case of water bottles? Remember you can use your Desmos Calculator
(1 point)
0.5 m/s2
2 m/s2
2.32 m/s2
2.7 m/s2
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All Answers 1
Answered by
GPT-5 mini
AI
I don’t have the full problem statement or the numbers (mass, forces, angle, friction coefficient, etc.), so I can’t pick the correct choice from those four. Can you paste the full problem or a picture?
Meanwhile, here are the formulas and steps you can use (and enter into Desmos) to find the acceleration depending on the situation:
1. If you know the net force Fnet on the case and its mass m:
a = Fnet / m
In Desmos: a = Fnet / m
2. On an incline without friction:
a = g * sin(θ)
In Desmos: a = 9.8 * sin(theta) (enter theta in radians or use sin(30 deg) syntax)
3. On an incline with kinetic friction μk:
a = g * (sin(θ) − μk * cos(θ))
In Desmos: a = 9.8*(sin(theta) - mu_k*cos(theta))
4. Horizontal push with friction:
a = (F_push − μk * m * g) / m = F_push/m − μk * g
In Desmos: a = Fpush/m - mu_k*9.8
If you post the numbers from your worksheet (mass, forces, angle, μ), I’ll compute the acceleration and pick which of the four choices is correct.
Meanwhile, here are the formulas and steps you can use (and enter into Desmos) to find the acceleration depending on the situation:
1. If you know the net force Fnet on the case and its mass m:
a = Fnet / m
In Desmos: a = Fnet / m
2. On an incline without friction:
a = g * sin(θ)
In Desmos: a = 9.8 * sin(theta) (enter theta in radians or use sin(30 deg) syntax)
3. On an incline with kinetic friction μk:
a = g * (sin(θ) − μk * cos(θ))
In Desmos: a = 9.8*(sin(theta) - mu_k*cos(theta))
4. Horizontal push with friction:
a = (F_push − μk * m * g) / m = F_push/m − μk * g
In Desmos: a = Fpush/m - mu_k*9.8
If you post the numbers from your worksheet (mass, forces, angle, μ), I’ll compute the acceleration and pick which of the four choices is correct.
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