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azamat
2 years ago
15

What should we do to be a computer engineer ‍ ‍?​

Physics
2 answers:
ycow [4]2 years ago
6 0

Answer:

to be an eginere u would have to go to college and study hard

Explanation:

____ [38]2 years ago
5 0
Study and go to university/college
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As an object falls freely near the surface of the earth, its velocity?
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If it were possible for an object to fall freely near the surface of the Earth,

-- The direction of its velocity would always be "down"; that is, toward the center of the Earth.

-- The size of its velocity would continually increase, at the rate of 9.8 meters per second for every second it falls.

7 0
3 years ago
A particle, of mass 6 kg, is in equilibrium on a rough horizontal plane under a force o-f magnitude T N, which acts at an angle
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Answer:

T is less than or equal to 19 N

Explanation:

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Temperature is the energy
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3 years ago
A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi
krok68 [10]

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

    (m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

                 k = [(m)×(v^2)]/(x^2)

                 k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

                 k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

             v^2 = [(k)(x^2)]/m

                 v =  \sqrt{ [(k)(x^2)]/m}

                 v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

                    = 0.491 m/s

8 0
2 years ago
Read 2 more answers
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