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swat32
3 years ago
15

For a mass oscillating on a spring at what positions are (a) velocity and (b) acceleration of the mass have maximum valeus?

Physics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

a)At the mean position

b)At the extremes positions

Explanation:

Given that mass is having oscillation motion.

We know that

1. At the mean position -The velocity of the mass is maximum and the acceleration of the mass is minimum.The net force on the mass will be zero.

2. At the extreme position-The velocity of the mass is minimum and the acceleration of the mass is maximum.The net force on the mass will not be zero.

Therefore

a)At the mean position

b)At the extremes positions

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Answer:

0.3677181864 m

Explanation:

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\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

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A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t
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Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

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