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sergey [27]
3 years ago
8

Explain two ways friction can be reduced

Physics
1 answer:
BlackZzzverrR [31]3 years ago
6 0
Friction can be reduced by making a surface smooth. It can also be reduced by applying lubricants such as oil and grease
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The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?
Svetradugi [14.3K]

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.

3 0
3 years ago
I believe Newton's 1st law is
Bogdan [553]

Answer:

The first law states that if the net force is zero, then the velocity of the object is constant.

5 0
3 years ago
If a proton were released in the electric field above, what direction would it move?
Afina-wow [57]

Answer:

d because the proton would move towards the negative plate

Explanation:

5 0
3 years ago
What's the difference and similarity between O2 and 2 O ?
nasty-shy [4]
O2 refers to two oxygen atoms bonded together, while 2 O refers to two oxygen atoms that are not bonded to each other. They both have two oxygen atoms, but in O2, the oxygen atoms are bonded, while in 2O, the atoms are not.
4 0
3 years ago
The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt
nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0
3 years ago
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