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mr_godi [17]
3 years ago
12

9 the only force acting on a 2.0 kg canister that is moving in an xy plane has a magnitude of 5.0 n. the canister initially has

a velocity of 4.0 m/s in the positive x direction and some time later has a velocity of 6.0 m/s in the positive y direction. how much work is done on the canister by the 5.0 n force during this time?
Physics
1 answer:
-Dominant- [34]3 years ago
6 0
We don't need to know the magnitude of the force.

The canister's initial kinetic energy is (1/2) M V₀² = 16 joules.

Its final kinetic energy is  (1/2) M V₁² = 36 joules.

Its kinetic energy has grown by 20 joules.  The additional energy
must have come from the 5.0 N force, since that's the only force
acting on the canister.  
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An energy storage system based on a flywheel (a rotating disk) can store a maximum of 3.7 MJ when the flywheel is rotating at 16
Likurg_2 [28]

The moment of inertia of the flywheel is 2.63 kg-m^{2}

It is given that,

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1 year ago
This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa
Montano1993 [528]

Answer:

2.77287\times 10^{15}\ m

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n = Number of photons per second

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

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r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s

Photon intensity is given by

i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m

The distance is 2.77287\times 10^{15}\ m

3 0
3 years ago
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