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rewona [7]
3 years ago
11

A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the

following should be the technician’s NEXT step?
A. Implement the solution.
B. Document lessons learned.
C. Establish a plan of action.
D. Verify full system functionality.
Physics
1 answer:
vladimir1956 [14]3 years ago
3 0

Answer:

b)  Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

You might be interested in
Which formulas have been correctly rearranged to solve for radius? Check all that apply. r = GM central/v^2 r =fcm/v^2 r =ac/v^2
jek_recluse [69]

The orbital radius is: r=\frac{GM}{v^2}

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

\frac{GMm}{r^2}=m\frac{v^2}{r}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the satellite

r is the radius of the orbit

v is the speed of the satellite

Re-arranging the equation, we find:

\frac{GM}{r}=v^2\\r=\frac{GM}{v^2}

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0
3 years ago
Read 2 more answers
Find the mass of a flying discus that has a net force of 1.05 newtons and accelerates at 3.5 m/s^2
Ilya [14]
F=ma
m=1.05/3.5= 0.3kg
8 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
A cliff diver from the top of a 100 (m) cliff. He begins his dive by jumping up with a velocity of 5 (m/s) a. How
Nutka1998 [239]

Answer:

A.) 4.81 seconds

B.) 44.6 m/s

Explanation:

He begins his dive by jumping up with a velocity of 5 (m/s).

Let us first calculate the maximum height reached by using third equation of motion

V^2 = U^2 - 2gH

At maximum height, V = 0

0 = 5^5 - 2 × 9.8H

19.6H = 25

H = 25 /19.6

H = 1.28 m

The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.

Where height h = 1.28 + 100 = 101.28 m

h = Ut + 1/2gt^2

As the diver drop from maximum height, U = 0

101.28 = 1/2 × 9.8 × t^2

4.9t^2 = 101.28

t^2 = 101.28/4.9

t^2 = 20.669

t = sqrt ( 20.669)

t = 4.55s

As the diver jumped up, the time taken to reach the maximum height will be

Time = 1.28 / 5 = 0.256

The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds

B.) Velocity right before he hits the water will be

V^2 = U^2 + 2gH

But U = 0

V^2 = 2 × 9.8 × 101.28

V^2 = 1985.09

V = 44.6 m/s

6 0
4 years ago
A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determi
sattari [20]

Answer:

(a) t = 3.74 s

(b) H = 136.86 m

(c) Vₓ = 41.83 m/s,  Vy = 0 m/s

(d) ax = 0 m/s²,  ay = 9.8 m/s²

Explanation:

(a)

Time to reach maximum height by the projectile is given as:

t = V₀ Sinθ/g

where,

V₀ = Launching Speed = 55.6 m/s

Angle with Horizontal = θ = 41.2°

g = 9.8 m/s²

Therefore,

t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)

<u>t = 3.74 s</u>

<u></u>

(b)

Maximum height reached by projectile is:

H = V₀² Sin²θ/g

H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)

<u>H = 136.86 m</u>

<u></u>

(c)

Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:

Vₓ = V₀ₓ = V₀ Cos θ

Vₓ = (55.6 m/s)(Cos 41.2°)

<u>Vₓ = 41.83 m/s</u>

Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.

<u>Vy = 0 m/s</u>

<u></u>

(d)

Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.

<u>ax = 0 m/s²</u>

The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:

<u>ay = 9.8 m/s²</u>

3 0
3 years ago
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