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Lemur [1.5K]
4 years ago
6

Two balls of equal mass at the bottom of an incline are rolled upward without slipping at the same initial velocity. one ball is

solid and the other is a thin-walled hollow ball. which rolls higher up the incline before coming to a stop?
Physics
1 answer:
german4 years ago
3 0
The appropriate response would be the hollow ball.The purpose for it has more kinetic energy at the last (2/3 as opposed to 2/5) so it will have more potential energy (more height) at the best. Pondering it, more inertia implies its less impervious to change meaning the hollow ball with more inertia wont need to stop as fast.
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A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

W=22.5 Nm+44.96 Nm

W=67.46 Nm

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So we have

-\left(1.0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2a(5.0\,\mathrm m)

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