All categories

3 years ago

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

You might be interested in

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.

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

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 = $\frac{dE_x}{dE}$