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Dmitry [639]
3 years ago
15

A computer technician always touches the metal body of a computer before touching any of its electrical parts. why?

Physics
1 answer:
Nikolay [14]3 years ago
6 0

Answer:

Computer technician always touch the metal body parts of the computer before touch any electrical part of the computer because it will neutralized the body static charge.

The components of the computer are made up of the materials with the positive and negative charge electron and their electric effect basically neutralized the one another.

while touching any electrical hardware make sure that the electricity is off and discharge our body from any static charge. Also avoid wearing wool clothes when touching any electrical hardware parts as they also build static charge.  

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What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?
KonstantinChe [14]

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

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3 years ago
When the area over which pressure is applied is increased,
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Answer:

A. the pressure decrease

Explanation:

pressure decreases when the surface area over which a force is applied increases. pressure increases when the surface area over which force is applied decreases.

3 0
3 years ago
Evan drew a diagram to illustrate radiation
labwork [276]
The answer would be D, electromagnetic waves
8 0
3 years ago
Read 2 more answers
How does the role of blood protect the body
maksim [4K]

Well it transports white blood cells to infected areas.

4 0
3 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
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