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Vlada [557]
3 years ago
5

Engineers often build a prototype perfectly. It is rare that they have to improve their product.

Physics
2 answers:
nata0808 [166]3 years ago
7 0

Answer:

false

Explanation:

most prototypes are and perfect and they are build to find our mistakes

Vesnalui [34]3 years ago
6 0
False because they might fail in the first try but they have to keep trying till it’s nearly perfect
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A bag is dropped from a balloon that is 50m above the ground and rising at 15m/s. calculate
ANTONII [103]

Answer:

See the answers below

Explanation:

We can solve this problem using the principle of energy conservation. That is, the energy is conserved before and after dropping the bag.

For this case we have mechanical energy, which is the sum of the kinetic and potential energies.

E_{1}=E_{2}

where:

E_{1}=E_{k}+E_{p}\\E_{2}=E_{p}

Ek = kinetic energy [J] (units of Joules)

Ep = potential energy [J]

In the final Energy (2), there is only potential energy. since when the balloon reaches the maximum height its velocity is zero, that is, there is no kinetic energy.

A)

m*g*h+\frac{1}{2}*m*v^{2} =m*g*h_{1} \\9.81*50+0.5*(15)^{2}=9.81*h_{1}\\h_{1} = 61.46 [m]

B)

With the value calculated above we can find the acceleration of the balloon.

The distance traveled is the difference between the maximum height and 50 meters.

x = 61.46-50\\x = 11.46[m]

With the following equation of kinematics.

v_{f}^{2} =v_{o}^{2}+2*a*x

0 = 15^{2} +2*a*11.46\\a = - 9.816 [m/s^{2} ]

The negative sign indicates that the acceleration acts downward. That is, in the opposite direction to the movement.

We can use the following equation of kinematics to find the final velocity after 4 seconds.

v_{f}=v_{o}-a*t\\v_{f}=15-9.816*(4)\\v_{f}=-24.24 [m/s]

Now the distance:

v_{f}^{2} =v_{o}^{2}-2*a*x\\(24.24)^{2} =(15)^{2} -2*9.81*x\\x = 18.48 [m]\\x_{f}=50+18.48 = 68.48 [m]

c) Using the following equation of kinematics.

v_{f}=v_{o}-a*t\\0 = 15-9.81*t\\15=9.81*t\\t = 1.52 [s]

4 0
3 years ago
An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.2 m and charge 0.021 mC. Suppose th
igomit [66]

Answer:

T = 1.12 10⁻⁷ s

Explanation:

This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.

All the charge dq is at a distance r

           dE = k dq / r²

Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry

            cos θ =\frac{dE_x}{dE}

             dEₓ = dE cos θ

              cos θ = x / r

substituting

                dEₓ = k \frac{dq}{r^2 } \ \frac{x}{r}

                DEₓ = k dq x / r³

let's use the Pythagorean theorem to find the distance r

             r² = x² + a²

where a is the radius of the ring

we substitute

              dEₓ = k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq

we integrate

               ∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} }  ∫ dq

               Eₓ = k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}

In the exercise indicate that the electron is very central to the center of the ring

                x << a

                Eₓ = k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}

if we expand in a series

                  (\ 1+ (x/a)^2 \  )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2

we keep the first term if x<<a

                 Eₓ = \frac{ k Q}{a^3} \ x

the force is

                 F = q E

                 F = - \frac{kQ  }{a^3} \ x

this is a restoring force proportional to the displacement so the movement is simple harmonic,

                 F = m a

                 - \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }

                 \frac{d^2 x}{dt^2} = \frac{keQ}{ma^3}  \ x

the solution is of type

                  x = A cos (wt + Ф)

with angular velocity

                w² = \frac{keQ}{m a^3}

angular velocity and period are related

                 w = 2π/ T

             

we substitute

               4π² / T² = \frac{keQ}{m a^3}

                T = 2π  \sqrt{\frac{m a^3 }{keQ} }

let's calculate

                T = 2π \sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19}  \ 0.021  \ 10^{-3} }  }

                 T = 2π pi \sqrt{320.426 \ 10^{-18} }

                 T = 2π  17.9 10⁻⁹ s

                 T = 1.12 10⁻⁷ s

6 0
3 years ago
the law of blank of matter and energy states that matter cannot be made nor destroyed but can only be changed from one form to a
MissTica
Law of conservation is what ur referring to
4 0
4 years ago
If a person weighs 717 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that
Feliz [49]

<u>Answer:</u>

The acceleration due to gravity on that planet = 72.76 m/s^2

<u>Explanation:</u>

  Weight of person in earth = 717 N

  We know that weight = mg

   Acceleration due to gravity of earth = 9.8m/s^2

   So mass of person = 717/9.8 = 73.16 kg

   Mass of body is constant everywhere, so mass of person on the surface of a nearby planet = 73.16 kg

   Weight on the surface of a nearby planet = 5320 N = mg', where g' is the acceleration due to gravity value of the nearby planet.

     So   5320 = 73.16*g'

                   g' = 72.76 m/s^2

   The acceleration due to gravity on that planet = 72.76 m/s^2

8 0
4 years ago
Can you pleas help me!
Sergeeva-Olga [200]

Answer:

T

Explanation:

4 0
4 years ago
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