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IrinaVladis [17]
3 years ago
13

A fancart of mass 0.8 kg initially has a velocity of < 0.9, 0, 0 > m/s. Then the fan is turned on, and the air exerts a co

nstant force of < -0.3, 0, 0 > N on the cart for 1.5 seconds. What is the change in momentum of the fancart over this 1.5 second interval? = < , 0, 0 > kg
Physics
1 answer:
Dmitry [639]3 years ago
6 0

Answer:

<-0.45, 0, 0> kgm/s

Explanation:

The change in momentum would equal to the impulse generated by force F = <-0.3, 0, 0> over time duration of Δt = 1.5 second. The impulse is defined as the product of force F and the duration Δt

\Delta p = F \Delta t = (-0.3) * 1.5 = -0.45 kgm/s

So the change in momentum is <-0.45, 0, 0> kgm/s

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Energy = 7.83 x 10⁻¹⁹ J

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Explanation:

The energy of a photon in terms of wavelength can be calculated by the following formula:

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Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\

<u>Energy = 7.83 x 10⁻¹⁹ J</u>

<u></u>

Now, for λ = 300 nm = 3 x 10⁻⁷ m:

Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\

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