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shtirl [24]
2 years ago
7

A car of mass 750kg accelerates away from traffic lights. At the end of the first 100m it has reached a speed of 12 m/s. During

this time, its engine provides an average forward force of 780 N, and the average force of friction on the car is 240 N. Calculate the increase in the car's kinetic energy at the end of the first 100 m
Physics
2 answers:
xenn [34]2 years ago
5 0

Answer:

the answer according to me is 54000j

Explanation:

im not sure why the force and friction are given inthe question because the the formula of ke is 1/mv^2

aliina [53]2 years ago
3 0

Answer:

Hruhriehrvdi8d8edeuebueufbrhddvuszu

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Answer:

Neither.

Explanation:

When an electron is released from rest, in an uniform electric field, it will accelerate moving in a direction opposite to the field (as the field has the direction that it would take a positive test charge, and the electron carries a negative charge).

It will move towards a point  with a higher potential, so its kinetic energy will increase, while its potential energy will decrease:

⇒ ΔK + ΔU = 0 ⇒ ΔK = -ΔU = - (-e*ΔV)

As ΔV>0, we conclude that the electric potential energy decreases while the kinetic energy increases in the same proportion, in order to energy be conserved, in absence of non-conservative forces.

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Which statement is TRUE concerning the particle arrangement of a solid
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It's packed together
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The host says during the video that her interests during middle school were?
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Answer:

huh? What video?

Explanation:

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3 years ago
Consider light of wavelengths 400 nm (UV), 550 nm (green, visible), and 900 nm (infrared). What is the energy associated with a
allsm [11]

Answer:

Energy of UV light =4.95\times 10^{-19}j

Energy of green light =3.6\times 10^{-19}j

Energy of infrared light =2.2\times 10^{-19}j

Explanation:

We have given the wavelength of UV light = 400 nm =400\times 10^{-9}m , wavelength of green light = 550 nm and wavelength of infrared = 900 nm

Speed of light c=3\times 10^8m/sec

Plank's constant h=6.6\times 10^{-34}

Energy of the signal is given by E=h\nu =h\frac{c}{\lambda }

So energy of UV light E=\frac{6.6\times 10^{-34}\times3\times 10^8}{400\times 10^{-9}}=4.95\times 10^{-19}j

Energy of green light E=\frac{6.6\times 10^{-34}\times3\times 10^8}{550\times 10^{-9}}=3.6\times 10^{-19}j

Energy of infrared light E=\frac{6.6\times 10^{-34}\times3\times 10^8}{900\times 10^{-9}}=2.2\times 10^{-19}j

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2 years ago
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<h3><u>Answer;</u></h3>

<em>When a wave meets a boundary, the angle of the reflected wave is equal to the angle of the incident wave</em>

<h3><u>Explanation</u>;</h3>
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