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3 years ago

The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?

Physics

1 answer:

Svetradugi [14.3K]3 years ago

3 0

Answer:

The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.

Explanation:

This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.

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Which of the following is not an example of a polymer?

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Concrete is not a polymer which Nylon, and Kevlar are

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An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal

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Answer and Explanation:

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- Combine 50cm and 30cm to get 80 cm.

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2 years ago

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Convert 13.1 miles to feet. Using one step conversion

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3 years ago

Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70

IceJOKER [234]

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

$A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2$

here $\omega _{o}^{2}= k/m$ = (6.30)/(0.135) = 46.67 N/m kg

$F / mA$ = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

$\omega ^{2}= \omega _{o}^{2}\pm F/m$

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π = 8.53 /6.28 or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

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3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

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3 years ago