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3 years ago

Creating a prototype and troubleshooting are steps in the development of?

2 answers:

8 0

Your answer is a new experiment, since it's something that hasn't been tried before, or has, but resulted in with lots of errors.

Zigmanuir [339]3 years ago

7 0

Answer:

Im not the surest of what the answer is but it's not "a new expirement" for anyone thats wondering.

Explanation:

You might be interested in

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

3 years ago

You are driving a car with an automatic transmission; the shifter is mounted on the steering column. When you press down on the

castortr0y [4]

Answer:

From the narrative in the question, there seem to have been a break failure and the ordered step of response to this problem is to

1) Put on the hazard light to inform other road users of a problem or potential fault with your car and so they should continue their journey with caution.

2) Avoid pressing on the acceleration pedal as this might cause the car to gradually slow down due to friction and gravity

3)Try navigate the car to the service lane. This is the less busy lane where cars are sometimes parked briefly.

4) Continuously pump the breaks to try stop the car. Continuously pumping the breaks might just help you build enough pressure to stop the car because often time, there are some pressure left in the break.

5) At this point, the speed of the car should be relatively slow. So at this point, you could try apply the emergency hand break. Do not pull the emergency hand breaks if the car is on high speed. Doing this may cause the car to skid off the road.

8 0

3 years ago

A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate

Romashka [77]

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0

3 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$