3 years ago

What is the potential energy of a 2 kilogram potted plant that is on a 1 meter high plant stand?

Physics

1 answer:

Monica [59]3 years ago

5 0

PE=mgh
PE=2×9.8×1=19.6 J

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Answer:

Explanation:

Given

radius $r=0.227 m$

Charge on surface $Q=6.03\times 10^{-6} C$

Point Charge inside sphere $q=1.15\times 10^{-6} C$

Electric Field at $r=0.735 m$

Treating Surface charge as Point charge and applying Gauss law

$E_{total}A=\frac{q_{enclosed}}{\epsilon _0}$

where A=surface area up to distance r

$E_{total}=\frac{Q+q}{4\pi r^2}$

$E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}$

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A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

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Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

v=3.9 m/s

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