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Kitty [74]
2 years ago
5

Technician A says test lights are great for quick tests on non-computerized circuits. Technician B says you can use a test light

to check SRS circuits and computer circuits. Who is correct?
Physics
1 answer:
Tpy6a [65]2 years ago
4 0

Answer:

that technician A is right

Explanation:

The test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits, these two values ​​are high and can light the bulb. In digital circuits the current is very small in the order of milliamps, so there is not enough power to turn on these lights.

From the above it is seen that technician A is right

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The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Elodia [21]

Explanation:

Below is an attachment containing the solution.

7 0
2 years ago
The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve
Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

\varepsilon _t=\dfrac{D-d}{d}

\varepsilon _t=\dfrac{32-30}{30}

\varepsilon _t=0.0667

We know that

n=-\dfrac{\epsilon _t}{\varepsilon _{long}}

\varepsilon _{long}=-\dfrac{\epsilon _t}{n}

\varepsilon _{long}=-\dfrac{0.0667}{0.45}

\varepsilon _{long}=-0.148

So the axial pressure

P=E\times \varepsilon _{long}

P=5\times 0.148

P=740 KPa

The movement in the sleeve

\Delta =\varepsilon _{long}\times L

\Delta =0.148\times 50

Δ=7.4 mm

6 0
1 year ago
jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a
Morgarella [4.7K]
Centripetal acceleration = (speed)² / (radius) .

Force = (mass) · (acceleration)

Centripetal force = (mass) · (speed)² / (radius) .

                             = (11 kg) · (3.5 m/s)² / (0.6 m)

                             = (11 kg) · (12.25 m²/s²) / (0.6 m)

                             =  (11 · 12.25) / 0.6  kg-m/s²

                             =      224.58 newtons.    (about 50.5 pounds)

That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by.  It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second.  If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.
6 0
2 years ago
Read 2 more answers
Mickey, a daredevil mouse of mass m , m, is attempting to become the world's first "mouse cannonball." He is loaded into a sprin
Sati [7]

Answer:

  h = v₀² / 2g ,      h = k/4g     x²

Explanation:

In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches

Starting point. Lower compressed spring

              Em₀ = K = ½ m v²

Final point. Highest on the path

             Em_{f} = U = mg h

             

As or no friction the energy is conserved  

              Em₀ =  Em_{f}

              ½ m v₀²² = m g h

             h = v₀² / 2g

We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse

                  ½ k x² = 2 g h

                  h = k/4g     x²

8 0
2 years ago
Read 2 more answers
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur
GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  \mu mg  =  \frac{mv^2}{r}

    where \mu is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0
2 years ago
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