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3 years ago

In order to increase the amount of work completed it is necessary to

Physics

2 answers:

Dimas [21]3 years ago

5 0

Increase the input energy to the system

navik [9.2K]3 years ago

3 0

Do all you can in one big day that you have time off or work on one thing then work on the other at the same time

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África tiene más de 1000 lenguas, pero solo 50 son habladas por más de 500 mil personas, explica si esto es un aspecto positivo

Misha Larkins [42]

Answer:

put it in English please

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2 years ago

PLEASEE HELP!!! LOTS OF POINTS AND BRAINLIEST!!!!!!!!

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2 years ago

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An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$

Therefore, the mass of the object is 5.045 lbm.

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2 years ago

Questions 15 out of 20

Olin [163]

Answer:

Explanation:

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3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

$E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

where

$\phi =$ Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

$\lambda =$ wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

$E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

Magnetic field Equation

$B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

Where B₀= E₀/c

$B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T$

$B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

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3 years ago