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Serggg [28]
3 years ago
10

Which practice will help prevent data loss in case of a computer malfunction?

Physics
2 answers:
Gnoma [55]3 years ago
8 0
You should back-up your computer every few weeks (or as often as needed).

As soon as you suspect that your computer (or any other backup-able device) is too slow or "glitch-y" you should back-up your device before you lose any important files or data.

PREVENTING MALFUNCTION:
→ Always set your computer on a hard & flat surface.
→ Avoid downloading programs from unknown or non-trusted sources.
→ Avoid using your computer in a hot area.
→ Consider installing a virus/malware protection software on your computer.

Vedmedyk [2.9K]3 years ago
3 0

Schedule and perform regular backups of your data. Installing antivirus software and updating it regularly is a good practice. You must also back up your data on a regular basis. This helps to prevent data losses due to unforeseen events. Such events may include a virus threat or computer malfunction.

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A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the
____ [38]
 <span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal 

it can cross and reach top of trajectory if its top height h = 1.5m 
and horizontal distance d = (1/2) Range 
--------------------------------------... 
let t be top height time 
at top height, vertical component of its velocity =0 
vy = 0 = u sin p - gt 
t = u sin p/g 
h = [u sin p]*t - 0.5 g[t[^2 
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g 
u^2 sin^2 p/2g = 1.5 
u^2 sin^2 p = 1.5*2*9.8 = 29.4 
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component 
===================== 
t = HALF the time of flight 
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g] 
1 = u^2 sin p cos p/g 
u sin p * u cos p = 9.8 
5.42 * u cos p = 9.8 
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component 
check>> 
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s 
u < less than fish's potential jump speed 6.26 m/s 

so it will able to cross</span>
8 0
2 years ago
After striking both mirrors, at what angle relative to the incoming ray does the outgoing ray emerge?
PIT_PIT [208]
The appropriate response is Zero degrees. The beam will leave the two mirrors along a way parallel to the one it came in on. This is the guideline of the corner reflector, which is frequently utilized as a radar target. Take note of that the corner reflector utilizes three reflecting surfaces (that are set up at 90o from each other) rather than the two like are being utilized here. Wikipedia has a truly awesome drawing that shows this two-dimentional issue pleasantly. A moment connection is given to the article on the corner reflector and the 3-D angles.
4 0
2 years ago
Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins
irga5000 [103]

Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

X(t) = 4t^2 - 3t + 40 ...... (1)

For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

But dy/dt = velocity.

If dy/dt = 0, then

60t - 15 = 0

60t = 15

t = 15/60

t = 0.25s

Substitutes t in equation (1)

Total displacement will be

X(t) = 4(0.25)^2 - 3(0.25) + 40

X(t) = 0.25 - 0.75 + 40

Total displacement = 39.5 m

To calculate instantaneous velocity, substitute t into equation (2)

V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

4 0
2 years ago
Brycen can cover half a basketball court (about 14 meters) in 4.0 seconds flat! How fast can he run?
Vedmedyk [2.9K]

Speed is equal to distance traveled divided by the time. So it's 3.5 m/s

6 0
3 years ago
Please Help On These 2 Questions!!!!! I severely need help!!!!!!
lys-0071 [83]
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?

4a. If the magnetic field lines are equally spaced apart, in other words share the same density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density) 

4b. The electric potential is measured in Volts(V) and is uniform along  the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.

I really hope this makes sense to you and that my pictures help! :)

3 0
3 years ago
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